LIBRARY 

OF  THE 

UNIVERSITY  OF  CALIFORNIA. 
GIFT  - 

Class 


PLANE    GEOMETRY 


BY 

C.  A.  HART 

i  \ 

INSTRUCTOR   OF   MATHEMATICS,    WADLEIGH    HIGH    SCHOOL,    NEW    YORK   CITY 

AND 

DANIEL    D.    FELDMAN 

HEAD   OF  DEPARTMENT   OF    MATHEMATICS,    ERASMUS   HALL   HIGH    SCHOOL,    BROOKLYN 


WITH   THE    EDITORIAL   COOPERATION   OF 
J.  H.  TANNER  AND  VIRGIL  SNYDER 

PROFESSORS    OF    MATHEMATICS    IN    CORNELL    UNIVERSITY 


, 


NEW   YORK  •:•  CINCINNATI  •:•  CHICAGO 

AMERICAN     BOOK     COMPANY 


oo i  7  I*" 

GIFT 

COPYRIGHT,  1911,  BY 

AMERICAN   BOOK   COMPANY. 

ENTERED  AT  STATIONERS'  HALL,  LONDON. 


1I.-F.    PLANE   GEOMETRY. 
W.  P.  I 


PREFACE 

THIS  book  is  the  outgrowth  of  an  experience  of  many  years 
in  the  teaching  of  mathematics  in  secondary  schools.  The 
text  has  been  used  by  many  different  teachers,  in  classes  of  all 
stages  of  development,  and  under  varying  conditions  of  sec- 
ondary school  teaching.  The  proofs  have  had  the  benefit  of 
the  criticisms  of  hundreds  of  experienced  teachers  of  mathe- 
matics throughout  the  country.  The  book  in  its  present  form 
is  therefore  the  combined  product  of  experience,  classroom  test, 
and  severe  criticism. 

The  following  are  some  of  the  leading  features  of  the  book  : 

TJie  student  is  rapidly  initiated  into  the  subject.  Definitions 
are  given  only  as  needed. 

The  selection  and  arrangement  of  theorems  is  such  as  to  meet 
the  general  demand  of  teachers,  as  expressed  through  the  Mathe- 
matical Associations  of  the  country. 

Most  of  the  proofs  have  been  given  in  full.  Proofs  of  some  of 
the  easier  theorems  and  constructions  are  left  as  exercises  for 
the  student,  or  are  given  in  an  incomplete  form ;  but  in  every 
case  in  which  the  proof  is  not  complete,  the  incompleteness  is 
specifically  stated.  The  authors  believe  that  the  proofs  of 
most  of  the  propositions  should  be  complete,  first,  in  order  to 
serve  as  models  for  the  handling  of  exercises ;  second,  to  pre- 
vent the  serious  error  of  making  the  student  feel  contented 
with  loose  and  slipshod  reasoning  which  defeats  the  main  pur- 
pose of  instruction  in  geometry ;  and  third,  as  an  excellent 
means  of  reviewing  the  previous  theorems  on  which  they 
depend. 

Tlie  indirect  method  of  proof  is  consistently  applied.  The 
usual  method  of  proving  such  propositions  as  Arts.  189  and 

iii 

223707 


IV  PREFACE 

415,  e.g.,  is  confusing  to  the  student.  The  method  used  here 
is  convincing  and  clear. 

The  exercises  are  carefully  selected.  In  choosing  the  exercises, 
each  of  the  following  groups  has  been  given  due  impor- 
tance : 

(a)  Concrete  exercises,  including  numerical  problems  and 
problems  of  construction. 

(6)  So-called  practical  problems,  such  as  indirect  measure- 
ments of  heights  and  distances  by  means  of  equal  and  similar 
triangles,  drawing  to  scale  as  an  application  of  similar  figures, 
problems  from  physics,  from  design,  etc. 

(c)  The  traditional  exercises  given  in  a  more  or  less  abstract 
setting. 

The  arrangement  of  the  exercises  is  pedagogical.  Exercises 
of  a  rather  easy  nature  are  placed  immediately  after  the  theo- 
rems of  which  they  are  applications,  instead  of  being  grouped 
together  without  regard  to  the  principles  involved  in  them. 
In  many  instances  the  exercises  are  so  arranged  as  to  consti- 
tute a  careful  line  of  development,  leading  gradually  from  a 
very  simple  construction  or  exercise  to  others  that  are  more 
difficult.  For  the  benefit  of  the  brighter  pupils,  however,  and 
for  review  classes,  large  lists  of  more  or  less  difficult  exercises 
are  grouped  at  the  end  of  each  book. 

TJie  definitions  of  plane  closed  figures  are  unique.  The  stu- 
dent's natural  conception  of  a  plane  closed  figure  is  not  the 
boundary  line  only,  nor  the  plane  only,  but  the  whole  figure 
composed  of  the  boundary  line  and  the  plane  bounded.  All 
definitions  of  closed  figures  involve  this  idea,  which  is  entirely 
consistent  with  the  higher  mathematics. 

The  numerical  treatment  of  magnitudes  is  explicit,  the  funda- 
mental principles  being  definitely  assumed  (Art.  336,  proof  in 
Appendix,  Art.  595).  This  procedure  is  novel  and  is  believed 
to  be  the  only  logical,  and  at  the  same  time  teachable,  method 
of  dealing  with  incommensurables.  Teachers  who  find  these 
subjects  too  difficult,  however,  can  easily  omit  them  without 
interruption  of  sequence, 


PREFACE  v 

The  area  of  a  rectangle  is  introduced  by  actually  measuring  it, 
thereby  obtaining  its  measure-number.  This  method  permits 
the  same  order  of  theorems  and  corollaries  as  is  used  in  the 
parallelogram  and  triangle.  The  correlation  with  arithmetic 
in  this  connection  is  valuable.  The  number  concepts  already 
found  so  useful  and  practical  in  the  modern  treatment  of  ratio 
and  proportion  have  been  developed  in  connection  with  areas, 
as  well  as  in  other  portions  of  the  book. 

Proofs  of  the  superposition  theorems  and  the  concurrent  line 
theorems  will  be  found  exceptionally  accurate  and  complete. 

The  many  historical  notes  are  such  as  will  add  life  and  interest 
to  the  work. 

The  carefully  arranged  summaries  throughout  the  book,  and 
the  collection  of  formulas  of  plane  geometry  at  the  end  of  the 
book,  it  is  hoped,  will  be  found  helpful  to  teacher  and  student 
alike. 

Argument  and  reasons  are  arranged  in  parallel  form.  This 
arrangement  gives  a  definite  model  for  proving  exercises,  ren- 
ders the  careless  omission  of  the  reasons  in  a  demonstration 
impossible,  leads  to  accurate  thinking,  and  greatly  lightens  the 
labor  of  reading  papers. 

Every  construction  figure  contains  all  necessary  construction 
lines.  This  method  keeps  constantly  before  the  student  a  model 
for  his  construction  work,  and  distinguishes  between  a  figure 
for  a  construction  and  a  figure  for  a  theorem. 

The  mechanical  arrangement  is  such  as  to  give  the  student 
every  possible  aid  in  comprehending  the  subject  matter. 

The  grateful  acknowledgment  of  the  authors  is  due  to  many 
friends  for  helpful  suggestions  ;  especially  to  Miss  Grace  A. 
Bruce  of  the  Wadleigh  High  School,  New  York  City;  to  Mr. 
Edward  B.  Parsons  of  the  Boys' High  School,  Brooklyn;  and 
to  Professor  McMahon  of  Cornell  University. 


CONTENTS 

PLANE   GEOMETRY 

PAGE 

SYMBOLS  AND  ABBREVIATIONS      .        .         .        .        .         .     viii 

INTRODUCTION       ... 1 

The  Subject  Matter  of  Geometry 1 

Four  Fundamental  Geometric  Concepts 2 

The  Four  Concepts  in  Reverse  Order 2 

Definitions  and  Assumptions 3 

Lines 4 

Use  of  Instruments 6 

Surfaces 7 

Angles 8 

. 
Assumptions        ..........       11 

Demonstrations  .         .         . 13 

BOOK  I.     RECTILINEAR  FIGURES 19 

Necessity  for  Proof      .........       21 

Polygons.     Triangles 22 

Superposition .25 

Measurement  of  Distances  by  Means  of  Triangles       ...       32 
Summary  of  Equal  Triangle  Theorems        .....       35 

Loci 42 

Summary  of  Unequal  Angle  Theorems        .         .  .         .64 

Summary  of  Unequal  Line  Theorems          .        .         .        .         .64 

Parallel  Lines 65 

Summary  of  Parallel  Line  Theorems 73 

Quadrilaterals.     Parallelograms 84 

Concurrent  Line  Theorems          .......     100 

Construction  of  Triangles 106 

Directions  for  the  Solution  of  Exercises      .         .        .    '     .         .110 
Miscellaneous  Exercises      .        .        .        .        .        •        •        .111 

vi 


CONTENTS  vii 

PAGE 

BOOK  II.     THE  CIRCLE        .         .         .         .  .         .         .113 

Two  Circles 131 

Measurement 133 

Constants  and  Variables.     Limits       ......  137 

Miscellaneous  Exercises       .        .         .         .         .        .         .         .  156 

BOOK  III.     PROPORTION  AND  SIMILAR  FIGURES        .         .  1G1 

Similar  Polygons 176 

Summary  of  Similar  Triangle  Theorems 183  \S 

Drawing  to  Scale 188 

Miscellaneous  Exercises       .        .         ...         .        .         .        .  206 

BOOK  IV.     AREAS  OF  POLYGONS 209 

Transformation  of  Figures  ........  223 

Miscellaneous  Exercises .  239 

Exercises  of  Greater  Difficulty    .         .         .         .         .         .         .241 

BOOK  V.    REGULAR  POLYGONS.    MEASUREMENT  OF  THE 

CIRCLE 245 

Measurement  of  the  Circumference  and  of  the  Circle          .         .  258 

Miscellaneous  Exercises 274 

Miscellaneous  Exercises  on  Plane  Geometry       ....  276 

Formulas  of  Plane  Geometry       .......  282 

APPENDIX  TO  PLANE  GEOMETRY 284 

Maxima  and  Minima 284 

Variables  and  Limits.     Theorems       .        .         .         .        .         .291 

To  Every  Straight  Line  there  belongs  a  Measure-number  .         .  294 

Discussion  of  Common  Measure  of  Lines    .....  295 

An  Angle  can  be  bisected  by  only  One  Line       ....  296 

Note  on  Axioms ...         .        .        .         .         .        .         .         •  296 

INDEX  297 


SYMBOLS   AND   ABBREVIATIONS 


=  equals,  equal  to,  is  equal  to. 
=£  does  not  equal. 
>  greater  than,  is  greater  than. 
<  less  than,  is  less  than. 
—  equivalent,  equivalent  to,  is  equiva- 
lent to. 

~  similar,  similar  to,  is  similar  to. 
^  is  measured  by. 
_L  perpendicular,    perpendicular    to,    is 

perpendicular  to. 
Js  perpendiculars. 
||  parallel,  parallel  to,  is  parallel  to. 
|| s  parallels. 

.  .  .  and  so  on  (sign  of  continuation). 
v  since. 
.-.  therefore. 
^  arc  ;  AD,  arc  AB. 
O,  OJ  parallelogram,  parallelograms. 
O,  ©  circle,  circles. 
Z,  A  angle,  angles. 
A,  A  triangle,  triangles. 

Q.E.D.     Quod  erat  demonstrandum,  lohich  was  to  be  proved. 

Q.E.F.     Quod  erat  faciendum,  which  was  to  be  done. 

The  signs  +,  — ,  x  ,  •*•  have  the  same  meanings  as  in  algebra. 


rt. 

right. 

str. 

straight. 

ext. 

exterior. 

int. 

interior. 

alt. 

alternate. 

def. 

definition. 

ax. 

axiom. 

post. 

postulate. 

hyp. 

hypothesis. 

prop. 

proposition. 

prob. 

problem. 

th. 

theorem. 

cor. 

corollary. 

cons. 

construction. 

ex. 

exercise. 

%. 

figure. 

iden. 

identity. 

com  p. 

complementary. 

sup. 

supplementary. 

adj. 

adjacent. 

homol. 

homologous. 

viii 


PLANE  GEOMETRY 


INTRODUCTION 


1.  The  Subject  Matter  of  Geometry.  In  geometry,  although 
we  shall  continue  the  use  of  arithmetic  and  algebra,  our  main 
work  will  be  a  study  of  what  will  later  be  denned  (§  13)  as 
geometric  figures.  The  student  is  already  familiar  with  the 
physical  objects  about  him,  such  as  a  ball  or  a  block  of  wood. 
By  a  careful  study  of  the  following  exercise,  he  may  be  led  to 
see  the  relation  of  such  physical  solids  to  the  geometric  figures 
with  which  he  must  become  familiar. 

Exercise.  Look  at  a  block  of  wood  (or  a  chalk  box).  Has  it  weight  ? 
color  ?  taste  ?  shape  ?  size  ?  These  are  called  properties  of  the  solid. 
What  do  we  call  such  a  solid  ?  A  physical 
solid.  Can  you  think  of  the  properties  of  this" 
solid  apart  from  the  block  of  wood  ?  Imagine 
the  block  removed.  Can  you  imagine  the 
space  which  it  occupied  ?  What  name  would 
you  give  to  this  space  ?  A  geometric  solid. 
What  properties  has  it  that  the  block  possessed  ?  Shape  and  size. 
is  it  that  separates  this  geometric  solid  from  surrounding  space  ? 
thick  is  this  surface?  Ho'w  many  surfaces  has  the  block?  Where  do 
they  intersect  ?  How  many  intersections  are  there  ?  How  wide  are  the 
intersections  ?  how  long  ?  What  is  their  name  ?  They  are  lines.  Do 
these  lines  intersect  ?  where  ?  How  wide  are  these  intersections  ?  how 
thick  ?  how  long  ?  Can  you  say  where  this  one  is  and  so  distinguish  from 
where  that  one  is  ?  What  is  its  name  ?  It  is  a  point. 

If  you  move  the  block  through  space,  what  will  it  generate  as  it  moves  ? 
What  will  the  surfaces  of  the  block  generate  ?  all  of  them  ?  Can  you 
move  a  surface  so  that  it  will  not  generate  a  solid  ?  Yes,  by  moving  it 
along  itselji  What  will  the  edges  of  the  block  generate  ?  Can  you  move 
an  edge  so  that  it  will  not  generate  a  surface  ?  What  will  the  corners 
generate  ?  Can  you  move  a  point  so  that  it  will  not  generate  a  line  ? 


What 
How 


2  .  PLANE   GEOMETRY 

tfOUR   FUNDAMENTAL   GEOMETRIC   CONCEPTS 

2.  The   space   in  which  we   live,  although   boundless   and 
unlimited  in  extent,  may  be  thought  of  as  divided  h>to  parts. 
A  physical  solid  occupies  a  limited  portion  of   space.      The 
portion   of  space  occupied   by    a  physical   solid   is   called   a 
geometric  solid. 

3.  A  geometric  solid  has  length,  breadth,  and   thickness. 
It  may  also  be  divided  into  parts.     The  boundary  of  a  solid  is 
called  a  surface. 

4.  A  surface  is   no  part  of   a  solid.      It   has   length  and 
breadth,  but  no  thickness.     It  may  also  be  divided  into  parts. 
The  boundary  of  a  surface  is  called  a  line. 

5.  A  line  is  no  part  of  a  surface.     It  has  length  only.     It 
may  also  be  divided  into  parts.     The  boundary  or  extremity 
of  a  line  is  called  a  point. 

A  point  is  no  part  of  a  line.  It  has  neither  length,  nor 
breadth,  nor  thickness.  It  cannot  be  divided  into  parts-  It  is 
position  only. 

THE  FOUR  CONCEPTS  IN  REVERSE  ORDER 

6.  As  we  have  considered  geometric  solid  independently  of 
surface,  line,  and  point,  so  we  may  consider   point  indepen- 
dently, and  from  it  build  up  to  the  solid. 

A  small  dot  made  with  a  sharp  pencil  on  a  sheet  of  paper 
represents  approximately  a  geometric  point. 

7.  If  a  point  is  allowed  to  move  in  space,  the  path  in  which 
it  moves  will  be  a  line. 

A  piece  of  fine  wire,  or  a  line  drawn  on  paper  with  a  sharp 
pencil,  represents  approximately  a  geometric  line.  This,  how- 
ever fine  it  may  be,  has  some  thickness  and  is  not  therefore  an 
ideal  t  or  geometric,  line. 

.8.  If  a  line  is  allowed  tp  move  in  space,  its  path  in  general 
will  be  a  surface. 


INTRODUCTION  3 

9.   If  a  surface  is  allowed*  to  move  in  space,  its  path  in 
general  will  be  a  geometric  solid. 

10.  A  solid  has  threefold  extent  and  so  is  said  to  have  three 
dimensions ;  a  surface  has  twofold  extent  and  is  said  to  have 
two  dimensions ;  a  line  has  onefold  extent  or  one  dimension ; 
a  point  has  no  extent  and  has  therefore  no  dimensions. 

11.  The  following  may  be  used  as  working  definitions  of 
these  four  fundamental  concepts : 

A  geometric  solid  is  a  limited  portion  of  space. 
A  surface  is  that  which  bounds  a  solid  or  separates  it  from 
an  adjoining  solid  or  from  the  surrounding  space. 
A  line  is  that  which  has  length  only. 
A  point  is  position  only. 

DEFINITIONS   AND   ASSUMPTIONS 

12.  The  primary  object  of  elementary  geometry  is  to  deter- 
mine, by  a  definite  process  of  reasoning  that  will  be  introduced 
and  developed  later,  the  properties  of  geometric  figures.     In 
all  logical  arguments  of  this  kind,  just  as  in  a  debate,  certain 
fundamental   principles   are   agreed  upon  at  the   outset,  and 
upon  these  as  a  foundation  the  argument  is  built.     In  ele- 
mentary  geometry   these   fundamental    principles   are   called 
definitions  and  assumptions. 

The  assumptions  here  mentioned  are  divided  into  two  classes, 
axioms  and  postulates.  These,  as  well  as  the  definitions,  will 
be  given  throughout  the  book  as  occasion  for  them  arises. 

13.  Def.      A    geometric   figure    is   a  point,  line,  surface,  or 
solid,  or  a  combination  of  any  or  all  of  these. 

14.  Def.      Geometry   is   the   science    which   treats    of    the 
properties  of  geometric  figures. 

15.  Def.     A  postulate  may  be  defined  as  the  assumption  of 
the  possibility  of  performing  a  certain  geometric  operation. 

Before  giving  the  next  definition,  it  will  be  necessary  to 
introduce  a  postulate. 


4  PLANE   GEOMETRY 

16.  Transference   postulate.    •  Any  geometric  figure   may   be 
moved  from  one  position  to  another  without  change  of  size  or 
shape. 

17.  Def.     Two    geometric   figures    are    said   to    coincide   if, 
when  either  is  placed  upon  the  other,  each  point  of  one  lies 
upon  some  point  of  the  other. 

18.  Def.     Two   geometric  figures  are  equal  if  they  can  be 
made  to  coincide. 

19.  Def.     The  process  of  placing  one  figure  upon  another  so 
that  the  two  shall  coincide  is  called  superposition. 

This  is  an  imaginary  operation,  no  actual  movement  taking 
place. 

LINES 

20.  A  line  is  usually  designated  by  two  capital  letters,  as 
line  AB.     It  may  be  designated 

also    by    a    small    letter    placed     

somewhere  011  the  line,  as  line  a. 

21.  Straight  Lines.     In  §  7  we 
learned  that  a  piece  of  fine  wire 
or  a  line  drawn  on    a   sheet   of 

paper  represented  approximately  a  geometric  line.  So  also  a 
geometric  straight  line  may  be  represented  approximately  by 
a  string  stretched  taut  between  two  points,  or  by  the  line 
made  by  placing  a  ruler  (also  called  a  straightedge)  on  a  flat 
surface  and  drawing  a  sharp  pencil  along  its  edge. 

22.  Questions.     How  does  a  gardener  test  the  strai^htness  of  the 
edge  of  a  flower  bed  ?     How  does  he  get  his  plants  set  out  in  straight 
rows  ?     How  could  you  test  the  straightness  of  a  wire  ?     Can  you  think 
of  a  wire  not  stra;ght,  but  of  such  shape  that  you  could  cut  out  a  piece 
of  it  and  slip  it  along  the  wire  so  that  it  would  always  fit  ?     If  you  re- 
versed this  piece,  so  that  its  ends  changed  places,  would  it  still  fit  along 
the  entire  length  of  the  wire  ?     If  you  turned   it  over,  would   it  fit  ? 
Would  the  piece  cut  out  fit  under  these  various  conditions  if  the  wire 
were  straight  ? 


INTRODUCTION  5 

23.  JDef.     A  straight  line  is  a  line  such  that,  if  any  portion 
of  it  is  placed  with  its  ends  in  the  line,  the  entire  portion  so 
placed  will  lie  in  the  line,  however  it  may  be  applied. 

Thus,  if  AB  is  a  straight  line,  and  if  any  portion  of  AB,  as 
CD,  is  placed  on  any  other  part 

of    AS,    with    its    ends   in   AB,       4 £ 2 B 

every  point  of  CD  will  lie  in  AB.  FlG-  2- 

A  straight  line  is  called  also  a  right  line.     The  word  line, 
unqualified,  is  understood  to  mean  straight  line. 

24.  Straight   line  postulate.     A  straight  line  may  be  drawn 
from  any  one  point  to  any  other. 

25.  Draw  a   straight   line   AB.     Can   you   draw    a  second 
straight  line  from  A  to  B  ?     If  so,  where  will  every  point  of  the 
second  line  lie  (§  23)  ?     It  then  follows  that : 

Only  one  straight  line  can  be  drawn  between  two  points;  i.e.  a 
straight  line  is  determined  by  two  points. 

26.  Draw  two  straight  lines  AB  and  CD  intersecting  in  point 
P.     Show  that  AB  and  CD  cannot  have  a  second  point  in  com- 
mon (§  2-3).     It  then  follows  that: 

Two  intersecting  straight  lines  can  have  only  one  point  in  com- 
mon; i.e.  two  intersecting  straight  lines  determine  a  point. 

27.  Def.     A  limited  portion  of  a  straight  line  is  called  a  line 
segment,  or  simply  a  line,  or  a  segment.     Thus,  in  Fig.  2,  AC, 
CD,  and  DB  are  line  segments. 

28.  Def.     Two  line  segments  which  lie  in  the  same  straight 
line  are  said  to  be  collinear  segments. 

29.  Def.     A    curved    line    (or 
curve)    is  a  line  no   portion  of 
which  is  straight,  as  GH. 

30.  Def.      A  broken   line  is   a 
line  made  up  of  different  succes- 
sive straight  lines,  as  Kl.  FIG,  3, 


6  PLANE   GEOMETRY 

31.  Use  of  Instruments.  Only  two  instruments  are  permitted 
in  the  constructions  of  plane  geometry :  the  ruler  or  straight- 
edge for  drawing  a  straight  line,  already  spoken  of  in  §  21 ;  and 
the  compasses,  for  constructing  circles  or  arcs  of  circles,  and 
for  transferring  line  segments  from  one  position  to  another. 

Thus  to  add  two  lines, 
as  AB  and  CD,  draw, 
with  a  ruler,  a  straight 
line  OX.  Place  one  leg 
of  the  compasses  at  A 
and  the  other  at  B. 
Next  place  one  leg  at 
0  and  cut  off  segment 
OM  equal  to  AB.  In  a 
similar  manner  lay  off 

MN  equal  to  CD.     Then 

0  M  NX 

AB+CD  =  OM+  MN  =  ON.  FIG.  4. 

Show  how  to  subtract  AB  from  ON.     What  is  the  remainder? 


Ex.  1.     Can  a  straight  line  move  so  that  its  path  will  not  be  a  surface  ? 
If  so,  how  ? 

Ex.  2.     Can  a  curved  line  move  in  space  so  that  its  path  will  not  be  a 
surface  ?     If  so,  how  ? 

Ex.  3.     Can  a  broken  line  move  in  space  so  that  its  path  will  not  be  a 
surface  ? 

Ex.  4.     Draw  three  lines  as  AB,  CD,  and  ^ p 

EF.    Construct  the  sum  of  AB  and  CD  ;  of  AB 

and  EF-,  of  AB,  CD,  and  EF.  C D 

Ex.  5.     Construct  :    («)   the  difference  be-       E— F 

tween  AB  and  CD  ;  (6)  the  difference  between  FlG   5 

CD  and  EF;  (c)   the  difference  between  AB 

and  EF.    Add  the  results  obtained  for  (a)  and  (6)  and  see  whether  the 

sum  is  the  result  obtained  for  (c). 

Ex.  6.    Draw  a  line  twrce  as  long  as  AB  (the  sum  of  AB  and  AB); 
three  times  as  long  as  AB. 


INTRODUCTION  7 

SURFACES 

32.  Plane  Surface.     It  is  well  known  that  the  carpenter's 
straightedge  is  applied  to  surfaces  to  test  whether  they  are 
flat  and  even.     If,  no  matter  where  the  straightedge  is  placed 
on  the  surface,  it  always  fits,  the  surface  is  called  a  plane. 
Now  if  we  should  use  a  powerful  magnifier,  we  should  doubt- 
less discover  that  in  certain  places  the  straightedge  did  not 
exactly  fit  the  surface  on  which  it  was  placed.     A  sheet  of 
fine  plate  glass  more  nearly  approaches  the  ideal. 

33.  Questions.     Test  the  surface  of  the  blackboard  with  a  ruler  to 
see  whether  it  is  a  plane.     How  many  times  must  you  apply  the  ruler  ? 
Can  you  think  of  a  surface  such  that 

the  ruler  would  fit  in  some  positions 
(a  great  many)  but  not  in  all  ?  Can 
you  think  of  a  surface  not  plane  but 
such  that  a  piece  of  it  could  be  cut 
out  and  slipped  along  the  rest  so  that 
it  would  fit  ?  Would  it  fit  if  turned 
over  (inside  out)  ? 

34.  Def.     A  plane  surface  (or 

plane)  is  a  surface  of  unlimited  F 

extent  such  that  whatever  two 

of  its  points  are  taken,  a  straight  line  joining  them  will  lie 

wholly  in  the  surface. 

35.  Def.     A  curved  surface  is  a  surface  no  portion  of  which 
is  plane. 

36.  Def.     A  plane  figure  is  a  geometric  figure  all  of  whose 
points  lie  in  one  plane.     Plane  Geometry  treats  of  plane  figures. 

37.  Def.     A  rectilinear  figure  is  a  plane  figure  all  the  lines 
of  which  are  straight  lines. 


Ex.  7.     How  can  a  plane  move  in  space  so  that  its  path  will  not  t 
a  solid  ?  / 

Ex.  8.     Can  a  curved  surface  move  in  space  so  that  its  path  will  not 
be  a  solid  ?     If  so,  bow  ? 


8  PLANE   GEOMETRY 

ANGLES 

38.  Def.     An  angle  is  the   figure  formed   by  two  straight 
lines  which  diverge  from  a  point. 

The  point  is  the  vertex  of  the  angle  and  the  lines  are  its 
sides. 

39.  An  angle  may  be  designated  by  a  number  placed  within 
it,  as  angle  1  and  angle  2  in  Fig.  7,  and  angle  3  in  Fig.  8.     Or 


L  M  H 

FIG.  9. 

three  letters  may  be  used,  one  on  each  side  and  one  at  the 
vertex,  the  last  being  read  between  the  other  two;  thus  in  Fig.  7, 
angle  1  may  be  read  angle  ABC,  and  angle  2,  angle  CBD.  An 
angle  is  often  designated  also  by  the  single  letter  at  its  vertex, 
when  no  other  angle  has  the  same  vertex,  as  angle  F  in  Fig.  8. 

40.  Revolution  postulate.     A  straight  line  may  revolve  in  a, 
plane,  about  a  point  as  a  pivot,  and  when  it  does  revolve  contin- 
uously from  one  position  to  another,  it  passes  once  and  only  once 
through  every  intermediate  position. 

41.  A  clear  notion  of  the  magnitude  of  an  angle   may  be 
obtained  by  imagining  that  its  two  sides  were  at  first  collinear, 
and  that  one  of  them  has  revolved  about  a  point  common  to  the 
two.     Thus  in  Fig.  8.  we  may  imagine  FG  first  to  have  been  in 
the  position  FE  and  then  to  have  revolved  about  F  as  a  pivot 
to  the  position  FG. 

42.  Def.     Two  angles  are  adjacent  if  they  have  a  common 
vertex  and  a  common  side  which  lies  between  them ;  thus  in 
Fig.  7,  angle  1  and  angle  2  are  adjacent;  also  in  Fig.  9,  angle 
HMK  and  angle  KML  are  adjacent. 


INTRODUCTION 


9 


43.  Two  angles  are  added  by  placing  them  so  that  they  are 
adjacent.  Their  sum  is  the  angle  formed  by  the  two  sides  that 
are  not  common ;  thus  in  Fig.  10,  the  sum  of  angle  1  and  angle 
2  is  angle  ABC. 

C 


B 


FIG.  10. 


44.  The  difference  between  two  angles  is  found  by  placing 
them  so  that  they  have  a  vertex  and  a  side  in' common  but  with 
the  common  side  not  between  the  other  two.  If  the  other  two 
sides  then  happen  to  be  collineaci*,  the  difference  between  the 
angles  is  zero  and  the  angles  are  equal.  If  the  other  sides  are 
not  collinear,  the  angle  which  they  form  is  the  difference  between 
the  two  angles  compared ;  thus  in  Fig.  11,  the  difference  between 
angle  1  and  angle  2  is  angle  ABC. 

A 


FIG.  11. 

45.  Def.  If  one  straight  line  meets  another  so  as  to  make 
two  adjacent  angles  equal,  each  of 
these  angles  is  a  right  angle,  and  the 
lines  are  said  to  be  perpendicular  to 
each  other.  Thus,  if  DC  meets  AB 
so  that  angle  BCD  and  angle  DCA 
are  equal  angles,  each  is  a  right  angle, 
and  lines  An  and  CD  are  said  to  be  4  '  C 

perpendicular  to  each  other.  FIG.  12. 


10 


PLANE  GEOMETRY 


46.   Def.     If  two  lines  meet,  but  are  not  perpendicular  to 
each  other,  they  are  said  to  be  oblique  to  each  other. 


47.  Def.  An  acute  angle  is  an  angle 
that  is  less  than  a  right  angle ;  as  angle  1, 
Fig.  13. 


FIG.  13. 


FIG.  14. 


48.  Def.     An    obtuse  angle  is    an  angle 
that  is  greater  than  a  right  angle  and  less 
than  two  right  angles ;  as  angle  2,  Fig.  14. 

49.  Def.     A    reflex    angle    is    an    angle 
that  is  greater  than  two  right  angles  and 
less  than  four  right  angles;    as  angle  2, 
Fig.  15. 


50.  Note.      Two    lines   diverging    from    the 
same   point,    as  BA   and   BC,    Fig.    15,    always 
form  two  positive  angles,  as  the  acute  angle  1 
and  the  reflex  angle  2.     Angle  1  may  be  thought 
of  as  formed  by  the  revolution  of  a  line  counter- 
clockwise from  the  position  BA  to  the  position 

BC,  and  should  be  read  angle  ABC.  Angle  2  may  be  thought  of  as1 
formed  by  the  revolution  of  a  line  counter-clockwise  from  the  position  BG 
to  the  position  BA,  and  should  be  read' angle  CBA. 

51.  Def.     Acute,  obtuse,  and   reflex   angles  are   sometimes 
called  oblique  angles. 


Ex.  9.  (a)  In  Fig.  16,  if  angle  1  equals  angle 
2,  what  kind,  of  angles  are  they  ? 

(6)  Make  a  statement  with  regard  to  the  lines 
AB  and  CD. 

(c)  If  angle  3  does  not  equal  angle  4,  what 
kind  of  angles  are  they  ? 

(d)  Make  a  statement  with  regard  to  the  lines 
AB  and  CE. 


FIG. 


Ex.  10.  A  plumb  line  is  suspended  from  the  top  of  the  blackboard. 
What  kind  of  angles  does  it  make  with  a  horizontal  line  drawn  on  the 
blackboard  ?  with  a  line  on  the  blackboard  neither  horizontal  nor  vertical  ? 


INTRODUCTION  11 

Ex.  11.  Suppose  the  minute  hand  of  a  clock  is  at  twelve.  Where 
may  the  hour  hand  be  so  that  the  two  hands  make  with  each  other : 
(a)  an  acute  angle  ?  (6)  a  right  angle  ?  (c)  an  obtuse  angle,  ? 

Ex.  12.  Draw :  (a)  a  pair  of  adjacent  angles ;  (6)  a  pair  of  non- 
adjacent  angles. 

Ex.  13.  Draw  two  adjacent  angles  such  that :  (a)  each  is  an  acute 
angle  ;  (6)  each  is  a  right  angle ;  (c)  each  is  an  obtuse  angle  ;  (d)  one 
is  acute  and  the  other  right ;  (e)  one  is  acute  and  the  other  obtuse. 

Ex.  14.  In  Fig.  17,  angle  1  +  angle  2  =  ?  angle 
3  +  angle  4  =  ?  angle  BAD  +  angle  DAF  =  ?  angle 
DAF -  angle  3  =  ?  angle  2  +  angle  DAF-  angle  4 
=  ?  angle  4  +  angle  BAE  —  angle  1  =  ? 

Ex.  15.  Name  six  pairs  of  adjacent  angles  in 
Fig.  17. 

Ex.  16.  Draw  two  non-adjacent  angles  that 
have:  (a)  a  common  vertex;  (6)  a  common  side; 
(c)  a  common  vertex  and  a  common  side. 


52.  Def.     A  line  is  said  to  be  bisected  if  it 
is  divided  into  two  equal  parts. 

53.  Def.     The  bisector  of  an  angle  is  the  line  which  divides 
the  angle  into  two  equal  angles.* 


Ex.  17.  Draw  a  line  AB,  neither  horizontal  nor  vertical,  (a)  Draw 
freehand  a  line  perpendicular  to  AB  and  not  bisecting  it :_  (6)  a  line 
bisecting  AB  and  not  perpendicular  to  it. 

ASSUMPTIONS 

54.  1.  TJiings  equal  to  the  same  thing,  or  to  equal  things,  are 
equal  to  each  other. 

2.  If  equals  are  added  to  equals,  the  sums  are.  equal. 

3.  If  equals  are  subtracted  from  equals,  the  remainders  are 


4.  If  equals  are  added  to  unequals,  the  sums  are  unequal  in 
the  same  order. 

5.  If  equals  are  subtracted  from  unequals,  the  remainders  are 
unequal  in  the  same  order. 

*  The  proof  that  every  angle  has  but  one  bisector  will  be  found  in  the  Appendix,  §  599. 


12  PLANE   GEOMETRY 

6.  If  unequals  are  subtracted  from  equals,  the  remainders  are 
unequal  in  the  reverse  order. 

7.  (a)  If  equals  are  multiplied  by  equals,  the  products  are 
equal;  (b)   if  unequals  are  multiplied  by  equals,  the  products  are 
unequal  in  the  same  order. 

8.  (a)  If  equals  are  divided  by  equals,  the  quotients  are  equal; 
(b)  if  unequals  are  divided  by  equals,  the  quotients  are  unequal 
in  the  same  order. 

9.  If  unequals  are  added  to  unequals,  the  less  to  the  less  and 
the  greater  to  the  greater,  the  sums  are  unequal  in  the  same  order. 

10.  If  three  magnitudes  of  the  same  kind  are  so  related  that 
the  first  is  greater  than  the  second  and  the  second  greater  than  the 
third,  then  the  first  is  greater  than 'the.  third. 

11.  Tlie  whole  is  equal  to  the  sum  of  all  its  parts. 

12.  TJie  whole  is  greater  than  any  of  its  parts. 

13.  Like  powers  of  equal  numbers  are  equal,  and  like  roots  of 
equal  numbers  are  equal. 

14.  Transference   postulate.      Any  geometric  figure   may   be 
moved  from  one  position   to  another  without  change  of  size  or 
shape.     (See  §  16.) 

15.  Straight  line  postulate  I.     A  straight  line  may  be  drawn 
from  any  one  point  to  any  other.      (See  §  24.) 

16.  Straight  line  postulate  II.     A  line  segment  may  be  pro- 
longed indefinitely  at  either  end. 

17.  Revolution  postulate.     A  straight  line  may  revolve  in  a 
plane,  about  a  point  as  a  pivot,  and  when  it  does  revolve  continu- 
ously from  one  position  to  another,  it  passes  once  and  only  once 
through  every  intermediate  position.     (See  §  40.) 

55.  Assumptions  1-13  are  usually  called  axioms.  That  is, 
an  axiom  may  be  defined  as  a  statement  whose  truth  is  as- 
sumed.*  

Ex.  18.  Illustrate  the  first  five  assumptions  above  by  using  arithmeti- 
cal numbers  only. 

Ex.  19.  Illustrate  the  next  five  by  using  general  numbers  (letters) 
pnly. 


INTRODUCTION  13 

DEMONSTRATIONS 

56.  It  has  been  stated  (§  12)  that  the  fundamental  princi- 
ples agreed  upon  at  the  outset  as  forming  the  basis  of  the  logi- 
cal arguments  in  geometry  are  called  definitions,  axioms,  and 
postulates.     Every  new  proposition  advanced,  whether  it  is  a 
statement  of  a  truth  or  a  statement  of  something  to  be  per- 
formed, must  by  a  process  of  reasoning  be  shown  to  depend 
upon  these  fundamental  principles.     This  process  of  reasoning 
is  called  a  proof  or  demonstration.     After  the  truth  of  a  state- 
ment has  thus  been  established,   it  in  turn  may  be  used  to 
establish  new  truths. 

The  propositions  here  mentioned  are  divided  into  two  classes. 
theorems  and  problems. 

57.  Def.     A  theorem  is  a  statement  whose  truth  is  required 
to  be  proved  or  demonstrated.     For  example,  "  If  two  angles  of 
a  triangle  are  equal,  the  sides  opposite  are  equal"  is  a  theorem. 

There  are  two  parts  to  every  theorem  :  the  hypothesis,  or 
the  conditional  part ;  and  the  conclusion,  or  the  part  to  be 
proved.  In  the  theorem  just  quoted,  "  If  two  angles  of  a  tri- 
angle are  equal"  is  the  hypothesis;  and  "the  sides  opposite  are 
equal "  is  the  conclusion. 


Ex.  20.      Write  out  carefully  the  hypothesis  and  the  conclusion  of 
each  of  the  following  : 

(a)  If  you  do  your  duty  at  all  times,  you  will  be  rewarded. 

(&)  If  you  try  to  memorize  your  proofs,  you  will  never  learn  geometry. 

(c)  You  must  suffer  if  you  disobey  a  law  of  nature. 

(d)  Things  equal  to  the  same  thing  are  equal  to  each  other. 

(e)  All  right  angles  are  equal. 


58.  Def.     A  corollary  is  a  statement  of  a  truth  easily  de- 
duced from    another   truth.      Its    correctness,  like  that  of   a 
theorem,  must  be  proved. 

59.  Def.    A  problem,  in  general,  is  a  question  to  be  solved. 
As  applied  to  geometry,  problems  are  of  two  kinds,  namely, 
problems  of  construction  and  problems  of  computation. 


14  PLANE   GEOMETRY 

60.  From   the   revolution   postulate    (§   40)  and   from  the 
definition  of  a  perpendicular  (§  45)  we  may  deduce  the  follow- 
ing corollaries : 

61.  Cor.  I.    At  every  point  in  a  straight  line  there  exists 
a  perpendicular  to  the  line. 


Given    line   AB   and    point    C    in  j' 


To  prove   that  there  exists  a  J_  to  ,2-" 

AB  at  C,  as  CD.  •' 


-E 


AC  B 

Let  CE  (Fig.  18)  meet  AB  at  C,  so  that  FIG.  18. 

Z  1  <  Z  2.    Then  if  CE  is  revolved  about  C 

as  a  pivot  toward  position  CM,  Z  1  will  continuously  increase  and  Z  2 
will  continuously  decrease  (§  40)-  .-.  there  must  be  one  position  of  CE, 
as  (7Z>,  in  which  the  two  A  formed  with  AB  are  equal.  In  this  position 
CD  A.AB  (§  45). 

62.   Cor.  II.      At   every  point  in  a  straight  line  there 
exists  only  one  perpendicular  to  the  line. 

Given    CD  ±  AB  at  C}  i.e.  Z.  BCD  — 

To  prove     CD  the   only  _L   to   AB  \ 

at  C.  \ 


Let    CE  (Fig    19)   be  any  line  from   C    A  C  B 

other  than  CD.     Let  CE  fall  between  OD  FIG.  19. 

and    CA*    Then  /.BCE>Z.  BCD  ;   i.e.  > 

ZD(L4.     And  Z.ECA<^DCA  (§  54,  12).    .-.  ^.BOTand^CM  are  not 
equal  and  CE  is  not  _L  to  AB  (§  45). 

63.  §§61  and  62  may  be  combined  in  one  statement  as 
follows : 

At  every  point  in  a  straight  line  there  exists  one  and  only  one 
perpendicular  to  the  line. 

*  A  similar  proof  may  be  given  if  we  suppose  CE  to  fall  between  CE  and  CD, 

• 


INTRODUCTION 

64.   Cor.  m.     All  right  angles  are  equal. 
B  D 


15 


FIG.  20. 


Given   Z.AOB  and  Z  CPD,  any  two  rt.  A. 
To  prove    Z  AOB  =  Z  CPD. 


Place  Z  CPD  upon  Z  A  OB  so  that  point  P  shall  fall  upon  point  0,  and 
so  that  PC  shall  be  collinear  with  OA.  Then  PD  and  OB,  both  being 
_L  to  0^4  at  0,  must  be  collinear  (§  62).  .-.  the  two  A  coincide  and  are 
equal  (§  18). 

65.  Cor.  IV.  If  one  straight  line  meets  another  straight 
line,  the  sum  of  the  two  adjacent  angles  is  two  right  angles. 

Given  str.  line  CD  meeting  str.  line 
AP.  at  ct  forming  A  BCD  and  DC  A. 
To  prove  Z  BCD  +  Z  DC  A  =  2  rt.  A. 

Let  CE  be  J_  to  AB  at  C  (§  63). 
Then  Z  .BCD  +  Z  DCE  =  1  rt.  Z ; 

.-.  Z  BCD  =  1  rt.  Z  -/.DCE.       r— 
Again  Z  D(L4  =  1  rt.  Z  +  Z  D  C7.E. 

.-.  Z  BCD  +  Z  DC  A  =  2  rt.  4. 


c 

FIG.  21. 


66.  Cor.  V.  The  sum  of  all  the  angles  about  a  point  on 
one  side  of  a  straight  line  passing  through  that  point 
equals  two  right  angles. 

Given  A  1,  2,  3,  4,  5,  and  6, 
all  the  A  about  point  0  on  one 
side  of  str.  line  CA. 

To  prove    Z1+Z2  +  Z3+ 


The  sum  of  the  six  A  (Fig.22)  is  equal  to  /.AOB  +  Z  BOC. 


16 


PLANE   GEOMETRY 


67.   Cor.  VI.     The  sum  of  all  tKe  angles 
about  a  point  equals  four  right  angles. 

Prolong  one  of  the  lines  through  the  vertex  and  apply 
Cor.  V. 


— kl 


\ 

68.  Def.     Two  angles  whose  sum  is  one  right          FIG.  23. 
angle  are  called  complementary  angles,  as  angles  1  and  2,  Fig.  25. 
Either  of  two  such  angles  is  said  to  be  the  complement  of  the 
other. 

69.  Def.     Two  angles  whose   sum  is   two  right  angles  are 
called  supplementary  angles,  as  angles  1  and  2,  Fig.  26.     Either 
of  two  such  angles  is  said  to  be  the  supplement  of  the  other. 

An  angle  that  is  equal  to  the  sum  of  two  right  angles  is 
sometimes  called  a  straight  angle,  as  angle 


Fig.  26. 


70.  Def.  Two  angles  are  said  to  be 
vertical  if  the  sides  of  each  are  the  pro- 
longations of  the  sides  of  the  other; 
thus,  in  Fig.  24,  angles  1  and  2  are  ver- 
tical angles,  and  angles  3  and  4  are  like- 
wise vertical  angles. 


71.  Def.  An  angle  of  one  degree  is  one  nine- 
tieth of  a  right  angle.  A  right  angle,  therefore, 
contains  90  angle  degrees. 


FIG.  24. 


C 


Ex.  21.  In  Fig.  25,  angle  ABC  is  a  right  angle. 
If  angle  1  =  40°,  how  many  degrees  are  there  in  angle 
2  ?  How  many  degrees,  then,  are  there  in  the  comple- 
ment of  an  angle  of  40°  ? 


FIG.  25. 


Ex.  22.     How  many  degrees  are  there  in  the  complement  of  20°? 
35°  ?  of  a°  ?   of  |  right  angles  ?   of  k  right  angles  ? 

Ex.  23.     In  Fig.  26,  angle  1  +  angle  2,  or  angle 
ABC,  equals  2  right  angles.     If  angle   1  =  40°,      < 
how  many  degrees  are  there  in  angle  2  ?     How 


of 
D 


B 


FlG- 


.many  degrees,  then,  are  there  in  .the  supplement  of  .an  angle  of  40°  •? 


INTRODUCTION  17 

Ex.  24.  How  many  degrees  are  there  in  the  supplement  of  20°  ?  of 
140°-?  of  o°  ?  of  n  right  angles  ?  Give  these  answers  in  right  angles. 

Ex.  25.     In  the  accompanying  diagram  : 

(«)  If  angle  1  =  65°,  how  many  degrees  are  there  in  each  of  the  other 
three  angles  ? 

(6)  If  angle  2  =  ra°,  how  many  right  angles  are 
there  in  each  of  the  other  three  angles  ? 

(c)  If  angle  3  =  n  right  angles,  how  many  degrees 
are  there  in  each  of  the  other  three  angles  ? 

Ex.  26.     Compare  the  supplement  of  an  angle  of 

50°  with  its  complement.    *Draw  a  diagram  showing  the  complement  and 
the  supplement  of  an  acute  angle,  ABC. 

Ex.  27.  Criticize  the  following  exercise  :  How  many  degrees  are 
there  in  an  angle  whose  complement  is  f  of  its  supplement  ?  in  an  angle 
whose  supplement  is  f  of  its  complement  ? 

Ex.  28.  If  one  straight  line  meets  another  straight  line  so  that  one 
angle  is  double  its  adjacent  angle,  how  many  degrees  are  there  in  each  of 
the  two  adjacent  angles  ? 

Ex.  29.  How  many  degrees  are  there  in  an  angle  whose  complement 
and  supplement  together  equal  194°  ? 

Ex.  30.  How  many  degrees  are  there  in  an  angle  which  is  f  of  its 
complement? 

Ex.  31.  How  many  degrees  are  Inhere  in  an  angle  whose  supplement 
is  nineteen  times  its  complement  ? 

Ex.  32.  If  there  are  only  five  angles  about  a  point,  and  each  differs  , 
by  15°  from  an  angle  adjacent,  how  many  degrees  are  there  in  each  \J^ 
angle  ? 

Ex.  33.  If  there  are  only  six  angles  about  a  point  and  they  are  all 
equal,  how  large  is  each  angle  ? 


72.     Def.      Angles     that     are    -supple-      _  2  -s 
mentary  and   adjacent    are    called,  supple- 
mentary-adjacent angles,  as  A  l~and  2. 


Ex.  34.  If  two  angles  are  supplementary-adjacent  and  one  of  them  is 
one  half  a  right  angle,  how  large  is  the  other  ? 

Ex.  35.  Suppose  that  only  three  angles  are  formed  about  a  point  on 
one  side  of  a  straight  line  passing  through  the  point.  The  greatest  is  four 
times  the  least,  and  the  remaining  one  is  12°  more  than  the  least.  How 
many  degrees  are  there  in  each  ?  /) 


18  PLANE   GEOMETRY 

Ex.  36.      (a)  Can  two  angles  be  supplementary  and  not  adjacent? 
Illustrate. 

(6)  Can  two  angles  be  adjacent  and  not  supplementary  ?     Illustrate. 

(c)  Can  two  angles  be  both  supplementary  and  adjacent  ?     Illustrate. 

(d)  Can  two  angles  be  neither  supplementary  nor  adjacent  ?    Illustrate. 

(e)  Can  two  angles  be  both  complementary  and  adjacent  ?     Illustrate. 
(/)  Can  two  angles  be   both   complementary   and    supplementary  ? 

Illustrate. 


73.  From   the   definitions   of    complementary   and   supple- 
mentary angles  may  be  deduced  three  additional  corollaries  as 
follows : 

74.  Cor.  I.     Complements  of  the  same  angle  or  of  equal 
angles  are  equal. 

75.  Cor.  II.     Supplements  of  the  same  angle  or  of  equal 
angles  are  equal. 

76.  Cor.  III.     If  two   adjacent  angles  are   supplemen- 
tary, their  exterior  sides  are  collinear. 

Given    Z  A  EG  +  Z  CBD  =  2  rt.  A 
To  prove  BD  the  prolongation  of  AB. 

If   BD  is  not  the  prolongation  of  AB, 
then  some  other  line  from  B,  as  BE,  must     D  ~B  A 

be  its  prolongation.  FIG.  29. 

Then  Z  ABC  +  Z  CBD  =  2  rt.  A     (By  hyp.). 
Also  Z  ABC  +  Z  CBE  =  2  rt.  A     (§  65). 

This  is  impossible  (§  54,  12)  ;  i.e.  the  supposition  that  BE  is  the  pro- 
longation of  AB  leads,  by  correct  reasoning,  to  the  impossible  conclusion 
that  Z  CBD  =  Z  CBE.     Hence  this  supposition  itself  is  false. 
/.  BD  is  the  prolongation  of  AB. 


BOOK   I 

RECTILINEAR   FIGURES 

PROPOSITION   I.     THEOREM 

77.   If  two  straight  lines  intersect,  the  vertical  angles  are 
equal. 


Given  two  intersecting  str.  lines,  forming  the  vertical  A, 
and  3,  also  2  and  4. 

To  prove  Z 1  =  Z3  and  Z2  =  Z4. 


ARGUMENT 


=  2rt.  A 


4. 


5.   Likewise  Z  2  =  Z  4. 


Q.E.D. 


REASONS 

1.  If  one  str.  line  meets  an- 

other str.  line,  the  sum 
of  the  two  adj.  A  is  2 
rt.  Zs.  §  65. 

2.  Same  reason  as  1. 

3.  Things  equal  to  the  same 

thing  are  equal  to  each 
other.  §  54,  1. 

4.  If    equals    are    subtracted 

from  equals,  the  remain- 
ders are  equal.  §  54,  3. 

5.  By  steps  similar  to  1,  2,  3, 

and  4. 


78.  Cor.  If  two  straight  lines  intersect  so  that  any  two 
adjacent  angles  thus  formed  are  equal,  all  the  angles  are 
equal,  and  each  angle  is  a  right  angle. 

19 


20  PLANE   GEOMETRY 

Ex.  37.  If  three  straight  lines  intersect  at  a  common  point,  the  sum 
of  any  three  angles,  no  two  of  which  are  adjacent,  as  1,  3,  and  5,  in 
the  accompanying  diagram,  is  equal  to  two 
right  angles. 

Ex.  38.     In  the  same  diagram,  show  that : 

(a)  Angle  1  +  angle  3  =  angle  7. 

(&)  Angle  7  —  angle  4  =  angle  6. 

(c)  Angle  7  +  angle  4  —  angle  3  =  twice 
angle  1. 

(d)  Angle  2  -f  angle  4  +  angle  6  =  2  right  angles. 

Ex.  39.  The  line  which  bisects  one  of  two  vertical  angles  bisects  the 
other  also. 

Ex.  40.  The  bisectors  of  two  adjacent  complementary  angles  form 
an  angle  of  45°. 

Ex.  41.  Show  the  relation  of  the  bisectors  of  two  adjacent  supple- 
mentary angles  to  each  other. 

Ex.  42.  The  bisectors  of  two  pairs  of  vertical  angles  are  perpendicu- 
lar to  each  other. 

79.  The  student  will  observe  from  Prop.  I  that  the  complete 
solution  of  a  theorem  consists  of  four  distinct  steps : 

(1)  The  statement  of  the  theorem,  including  a  general  state- 
ment of  the  hypothesis  and  conclusion. 

(2)  The  drawing  of  a  figure  (as   general   as   possible),  to 
satisfy  the  conditions  set  forth  in  the  hypothesis. 

(3)  Tlie  application,  i.e.  the  restatement  of  the  hypothesis 
and  conclusion  as  applied  to  the  particular  figure  just  drawn, 
under  the  headings  of  "  Given  "  and  "  To  prove." 

(4)  The  proof,   or  demonstration,  the   argument  by   which 
the  truth  or  falsity  of  a  statement  is  established,  with  the 
reason  for  each  step  in  the  argument. 

80.  The  student  should  be  careful  to  quote  the  reason  for 
every  step   in  the  argument  which   he  makes  in   proving  a 
theorem.     The  only  reasons  admissible  are  of  two  kinds : 

1.  Something  assumed,  i.e.  definitions,  axioms,  postulates, 
and  the  hypothesis. 

2.  Something  previously  proved,  i.e.  theorems,  corollaries, 
and  problems. 


BOOK   I 


21 


81.  The  necessity  for  proof.  Some  theorems  seem  evident 
by  merely  looking  at  the  figure,  and  the  student  will  doubtless 
think  a  proof  unnecessary.  The  eye,  however,  cannot  always 
detect  error,  and  reasoning  enables  us  to  be  sure  of  our  conclu- 
sions. The  danger  of  trusting  the  eye  is  illustrated  in  the 
following  exercises.* 


Ex.  43.     In  the  diagrams  given  below,  tell  which  line  of  each  pair  is 
the  longer,  a  or  &,  and  verify  your  answer  by  careful  measurement. 


a 

r       n  — 
E6  —  I 

—1 

Ex.  44.     In  the  figures  below,  are  the  lines  everywhere  the  same  dis- 
tance apart  ?     Verify  your  answer  by  using  a  ruler  or  a  slip  of  paper. 


\   \    \   \   \    \   \    v 
\\\\\\\\ 


Ex.  45.     In  the  figures  below,  tell  which  lines  are  prolongations  of 
other  lines.     Verify  your  answers. 


*  These  diagrams  are  taken  by  permission  from  the  Report  of  the  Committee  on  Geom- 
etry, Proceedings  of  the  Eighth  Meeting  of  the  Central  Association  of  Science  an4 
Mathematics  Teachers, 


22  PLANE   GEOMETRY 

POLYGONS.     TRIANGLES 

82.  Def.     A  line  on  a  plane  is  said  to  be  closed  if  it  sepa- 
rates a  finite  portion  of  the  plane  from  the  remaining  portion. 

83.  Def.     A  plane  closed  figure  is  a  plane  figure  composed  of 
a  closed  line  and  the  finite  portion  of  the  plane  bounded  by  it. 

84.  Def.     A  polygon  is  a  plane  closed  figure  whose  boundary 
is  composed  of  straight  lines  only. 

The  points  of  intersection  of  the  lines  are  the  vertices  of 
the  polygon,  and  the  segments  of  the  boundary  lines  included 
between  adjacent  vertices  are  the  sides  of  the  polygon. 

85.  Def.     The  sum  of  the  sides  of  a  polygon  is  its  perimeter. 

86.  Def.     Any  angle  formed  by  two  consecutive  sides  and 
found  on  the  right  in  passing  clockwise  around  the  perimeter 
of  a  polygon  is   called  an  interior  angle  of  the  polygon,  or, 
for    brevity,    an    angle   of 

the  polygon.  In  Fig.  1, 
A  ABC,  BCD,  CDE,  DEA, 
and  EAB  are  interior 
angles  of  the  polygon. 

87.  Def.     If    any    side 
of  a  polygon  is  prolonged 
through  a  vertex,  the  angle 
formed   by   the    prolonga- 
tion and  the  adjacent  side  is  called  an  exterior  angle  of  the 
polygon. 

In  Fig.  1,  A  1,  2,  3,  4,  and  5  are  exterior  angles. 

88.  Def.     A  line  joining  any  two  non-adjacent  vertices  of  a 
polygon  is  called  a  diagonal;  as  AC,  Fig.  1. 

89.  Def.     A  polygon  which  has  all  of  its  sides  equal  is  an 
equilateral  polygon. 

90.  Def.     A  polygon  which  has  all  of  its  angles  equal  is  an 
equiangular  polygon. 


BOOK  I  23 

91.  Def.     A   polygon  'which  is  both  equilateral  and  equi- 
angular is  a  regular  polygon. 

92.  Def.     A  polygon  of  three  sides  is  called  a  triangle ;  one 
of  four  sides,  a  quadrilateral ;  one  of  five  sides,  a  pentagon ;  one 
of  six  sides,  a  hexagon ;  and  so  on. 

TRIANGLES    CLASSIFIED    WITH    RESPECT    TO    SIDES 

93.  Def.     A  triangle  having  no  two  sides  equal  is  a  scalene 
triangle. 

94.  Def.     A  triangle  having  two  sides  equal  is  an  isosceles 
triangle.     The  equal  sides  are  spoken  of  as  the  sides  *  of  the 
triangle.      The  angle   between  the  equal   sides  is  the  vertex 
angle,  and  the  side  opposite  the  vertex  angle  is  called  the  base. 

95.  Def.     A   triangle   having   its   three   sides  equal   is   an 
equilateral  triangle. 


Scalene  Isosceles  Equilateral 

TRIANGLES    CLASSIFIED    WITH    RESPECT    TO    ANGLES 

96.  Def.     A  right  triangle  is  a  triangle  which   has  a  right 
angle.     The  side  opposite  the  right  angle  is  called  the  hypote- 
nuse.    The  other  two  sides  are  spoken  of  as  the  sides  f  of  the 
triangle. 

97.  Def.     An   obtuse   triangle   is   a   triangle  which  has  an 
obtuse  angle. 

98.  Def.     An   acute  triangle  is  a  triangle  in  which  all  the 
angles  are  acute. 

*  The  equal  sides  are  sometimes  called  the  arms  of  the  isosceles  triangle.  This  term 
will  be  used  occasionally  in  the  exercises. 

t  Sometimes  the  sides  of  a  right  triangle  including  the  right  angle  are  called  the  arms 
of  the  triangle.  This  term  will  be  found  in  the  exercises. 


24  PLANE    GEOMETRY 

Ex   46.    Draw  a  scalene  triangle  freehand  :    («)  with  all  its  angles 
acute  and  with  its  shortest  side  horizontal ;  (&)  with  one  right  angle. 

Ex.  47.    Draw  an  isosceles  triangle  :  (a)  with  one  of  its  arms  hori- 
zontal and  one  of  its  angles  a  right  angle  ;  (6)  with  one  angle  obtuse. 


99.  Def.     The  side  upon  which  a  polygon  is  supposed   to 
stand  is  usually  called  its  base ;  however,  since  a  polygon  may 
be  supposed  to  stand  upon  any  one  of  its  sides,  any  side  may  be 
considered  as  its  base. 

The  angle  opposite  the  base  of  a  triangle  is  the  vertex  angle, 
and  the  vertex  of  the  angle  is  called  the  vertex  of  the  triangle. 

100.  Def.     The  altitude  of  a  triangle  is  the  perpendicular  to 
its  base  from  the  opposite  vertex.     In  general  any  side  of  a 
triangle  may  be  considered  as 

its  base.     Thus  in  triangle  EFG, 

if  FG  is  taken  as  base,  EH  is  the 

altitude;  if  GE  is  taken  as  base, 

^Jfwill  be  the  altitude;  if  EF 

is  taken  as  base,  the  third  alti-  e  „ 

tude  can  be  drawn.    Thus  every  FlG  A 

triangle  has  three  altitudes. 

It  will  be  proved  later  that  one  perpendicular,  and  only  one, 
can  be  drawn  from  a  point  to  a  line. 

101.  The  sides  of  a  triangle  are  often  designated  by  the 
small   letters   corresponding  to  the  capitals  at  the  opposite 
vertices ;  as,  sides  e,  /,  and  g,  Fig.  1. 


Ex.  48.  Draw  an  acute  triangle ;  draw  its  three  altitudes  freehand. 
Do  they  seem  to  meet  in  a  point  ?  Where  is  this  point  located  ? 

Ex.  49.  Draw  an  obtuse  triangle  ;  draw  its  three  altitudes  freehand. 
Do  they  meet  in  a  point  ?  Where  is  this  point  located  ? 

Ex.  50.    Where  do  the  three  altitudes  of  a  right  triangle  meet  ? 


102.   Def.     The  medians  of  a  triangle  are  the  lines  from  the 
vertices  of  the  triangle  to  the  mid-points  of  the  opposite  sides. 


BOOK  I  25 

103.  Superposition.  When  certain  parts  of  two  figures  are 
given  equal,  we  can  determine  by  a  process  of  pure  reason 
whether  the  two  figures  may  be  made  to  coincide. 

This  process  is  far  more  accurate  than  the  actual  transfer- 
ence of  figures,  for  we  are  free  from  physical  errors  such  as 
have  been  referred  to  in  §  81. 

Problem.     Given  line  AB  less  than  CD.     Apply  AB  to  CD. 

Solution.     Place  point  A  upon  point 

C.     Make  AB  collinear  with   CD  and     A B 

let  B  fall  toward  D.     Then  B  will  fall     £, ^ 

between  C  and  Z>,  because  AB  <  CD. 

Question.     Under  what  hypothesis  would  B  fall  on  D  ?  beyond  D  ? 

Problem.  Given  angle  ABC  less  than  angle  BST.  Apply  angle 
ABC  to  angle  BST. 

Solution.     Place  point  B  ~ 

upon  point  S  arid  make  BA 
collinear  with  SB. 

Then  BC  will  fall  between 

SB  and  ST,  because  Z  ABC  

<ZBST.  B  AS  R 


Ex.  51.  Solve  the  problem  above  by  first  making  BC  collinear  with 
ST.  Under  what  hypothesis  would  BA  fall  on  KB  ?  outside  of  angle 
BST?  within  angle  BST?  Illustrate  each  answer  by  a  diagram.  Can 
you  choose  where  BA  will  fall  after  you  have  put  BC  on  ST?  Could 
you  have  chosen  where  BA  should  fall  at  first  ? 


104.  Note.  In  applying  one  figure  to'  another,  always  begin  with  a 
line.  Place  one  end  of  it  on  one  end  of  another  line  and  make  the  two 
lines  collinear  on  the  same  side  of  the  point. 

Throughout  the  process,  determine  first  the  direction  each  line  will 
take,  and  then  where  its  end  will  fall. 

If  two  lines  are  given  equal,  one  maybe  placed  upon  the  other,  end  on 
end,  for  the  lines  will  coincide. 


Ex,  52.     Given  two  triangles  ABC  and  DEF,  such  that:    (1)  AC  = 
DF,  angle  A  is  greater  than  angle  Z),  angle  C  is  less  than  angle  F '; 

(2)  AC  =  DF,   angle   A  =  angle   D,    angle    C  is  less  than    angle   F; 

(3)  AC  =  DF,  angle  A  =  angle  Z>,  angle  C  =  angle  F.     Apply  triangle 
ABC  to  triangle  DEF  and.  draw  a  diagram  to  illustrate  each  case. 


26 


PLANE   GEOMETRY 


PROPOSITION  II.     THEOREM 

105.  Two  triangles  are  equal  if  a  side  and  the  two 
adjacent  angles  of  one  are  equal  respectively  to  a  side  and 
the  two  adjacent  angles  of  the  other. 


B 


Given  A  ABC  and  DEF,  AG  —  DF,  Z.  A  —  Z  D,  and  Z  (7=  Z  F. 
To  prove  A  ABC  =  A  DEF. 


ARGUMENT 

1.  Place  A  ABC  upon  A  DEF 

so  that  A  G  shall  fall  upon 
its  equal  DF,  A  upon  D, 
G  upon  F. 

2.  Then  .47?  will  become  col- 

linear  with  DEj  and  B 
will  fall  somewhere  on 
DE,  or  on  its  prolonga- 
tion. 

3.  Also  CB  will  become  col- 

linear  with  FE,  and  B 
will  fall  somewhere  on 
FE,  or  on  its  prolonga- 
tion. 

4.  .'.   point   B   must   fall   on 

point  E. 

5.  .'.  A  ABC  =  A  DEF. 

Q.E.D. 


REASONS 

1.  Any  geometric  figure  may 

be  moved  from  one  posi- 
tion to  another  without 
change  of  size  or  shape. 
§  54,  14. 

2.  Z  A  =  Z  D,  by  hyp. 


3.    Z  C=ZF,  by  hyp. 


4.  Two  intersecting  str.  lines 

determire  a  point.     §  26. 

5.  Two  geometric  figures  are 

equal    if    they    can    be 
made  to  coincide.     §  18. 


BOOK  I 


27 


Ex.  53.  In  the  figure  for  Prop.  II,  assume  AB  =  DE,  angle  A  — 
angle  D,  and  angle  B  =  angle  E\  repeat  the  proof  by  superposition, 
marking  the  lines  with  colored  crayon  as  soon  as  their  positions  are 
determined. 


B 


H 


D  6 


93° 


K 


E 


Ex.  54.  Place  polygon  I 
upon  polygon  II  so  that  some 
part  of  I  shall  fall  upon  its  equal 
in  II.  Discuss  the  resulting 
positions  of  the  remaining  parts 
of  the  figure. 

Ex.  55.  If  two  quadrilaterals  have  three  sides  and  the  included  angles 
of  one  equal  respectively  to  three  sides  and  the  included  angles  of  the 
other,  and  arranged  in  the  same  order,  are  the  quadrilaterals  equal? 
Prove. 

106.  Note.  The  method  of  superposition  should  be  used  for  proving 
fundamental  propositions  only.  In  proving  other  propositions  it  is  neces- 
sary to  show  merely  that  certain  conditions  are  present  and  to  quote 
theorems,  previously  proved,  which  state  conclusions  regarding  such 
conditions.  

Ex.  56.  If  at  any  point  in  the  bisector  of  an  angle  a  perpendicular  to 
the  bisector  is  drawn  meeting  the  sides  of  the  angle,  the  two  triangles 
thus  formed  will  be  equal. 

Ex.  57.     If  equal  segments,  measured  from  the  point  of  intersection 
of  two  lines,  are  laid  off  on  one  of  the  lines,  and  if  perpendiculars  to  this 
line  are  drawn  at  the  ends  of  these  segments,  two 
equal  triangles  will  be  formed. 

Ex.  58.  If  at  the  ends  of  a  straight  line  per-- 
pendiculars  to  it  are  drawn,  these  perpendiculars 
will  cut  off  equal  segments  upon  any  line  which 
bisects  the  given  line  and  is  not  perpendicular 
to  it. 


FIG.  1. 


FIG.  2. 


Ex.  59.  If  two  angles  of  a  triangle  are  equal,  the  bisectors  of  these 
angles  are  equal  (Fig.  1). 

Ex.  60.  If  two  triangles  are  equal,  the  bisector  of  any  angle  of  one  is 
equal  to  the  bisector  of  the  corresponding  angle  of  the  other  (Fig.  2). 


28 


PLANE   GEOMETRY 


PROPOSITION  III.     THEOREM 

107.  Two  triangles  are  equal  if  two  sides  and  the  in- 
cluded angle  of  one  are  equal  respectively  to  two  sides 
and  the  included  angle  of  the  other. 


Given    A  ABC  and  MNO,  AB  ==  MN,  AC  =  MO,  and  Z  A  = 
To  prove    A  ABC  =  A  MNO. 


M. 


ARGUMENT 

1.  Place  A  ABC  upon  A  MNO 

so  that  AC  shall  fall  upon 
its  equal  MO,  A  upon  3/> 
C  upon  0. 

2.  Then  AB  will  become  col- 

linear  with  MN. 

3.  Point  B  will  fall  on  point  N.- 

4.  . ' .  BC  will  coincide  with  NO. 


5.    .'.  A  ABC  =  A  MNO. 


Q.E.D. 


2. 


REASONS 

Any  geometric  figure  may 
be  moved  from  one  posi- 
tion to  another  without 
change  of  size  or  shape. 
§  54,  14. 

/.A  =  ZM,  by  hyp. 


3.  AB  =  MN,  by  hyp. 

4.  Only    one    str.     line     can 

be   drawn   between   two 
points.    §  25. 

5.  Two  geometric  figures  are 

equal  if  they  can  be  made 
to  coincide.    §  18. 


108.  Cor.  Two  right  triangles  are  equal  if  the  two  sides 
including  the  right  angle  of  one  are  equal  respectively 
to  the  two  sides  including  the  right  angle  of  the  other. 


E*.  61.    Prove  Prop.  Ill  by  placing  4B  upou  MN. 


BOOK  I  29 

109.  Del.     In  equal  figures,  the  points,  lines,  and  angles  in 
one  which,  when  superposed,  coincide  respectively  with  points, 
lines,  and   angles   in  the  other,  are  called  homologous  parts. 
Hence : 

110.  Homologous  parts  of  equal  figures  are  equal. 


Ex.  62.  If  two  straight  lines  bisect  each  other,  the  lines  joining  their 
extremities  are  equal  in  pairs. 

HINT.  To  prove  two  lines  or  two  angles  equal,  try  to  find  two  triangles, 
each  containing  one  of  the  lines  or  one  of  the  angles.  If  the  triangles  can 
be  proved  equal,  and  the  two  lines  or  two  angles  are  homologous  parts  of 
the  triangles,  then  the  lines  or  angles  are  equal.  The  parts  given  equal 
may  be  more  easily  remembered  by  marking  them  with  the  same  symbol, 
or  with  colored  crayon. 

Ex.  63.  In  case  the  lines  in  Ex.  62  are  perpendicular  to  each  other, 
what  additional  statement  can  you  make  ?  Prove  its  correctness. 

Ex.  64.  If  equal  segments  measured  from  the  vertex  are  laid  off  on 
the  sides  of  an  angle,  and  if  their  extremities  are  joined  to  any  point  in 
the  bisector  of  the  angle,  two  equal  triangles  will  be  formed. 

Ex.  65.  If  two  medians  of  a  triangle  are  perpendicular  to  the  sides 
to  which  they  are  drawn,  the  triangle  is  equilateral. 

Ex.  66.  If  equal  segments  measured  from  the  vertex  are  laid  off  on 
the  arms  of  an  isosceles  triangle,  the  lines  joining  the  ends  of  these 
segments  to  the  opposite  ends  of  the  base  will  be  equal  (Fig.  1). 


FIG.  1.  FIG.  2. 

Ex.  67.     Extend  Ex.  66  to  the  case  in  which   the   equal  segments 
are  laid  off  on  the  arms  prolonged  through  the  vertex  (Fig.  2). 


30 


PLANE   GEOMETRY 


PROPOSITION  IV.     THEOREM 

111.   The  base  angles  of  an  isosceles  triangle  are  equal. 

B 


Given   isosceles  A  ABC,  with  AB  and  BC  its  equal  sides. 
To  prove    Z  A  =  Z  C. 


ARGUMENT 

1.  Let  BD  bisect  Z  ABC. 

2.  In  A  ABD  and  DBC, 

AB  =  BC. 


BD  =  BD. 


A  ABD  =  A  DBC. 


6. 


Q.E.D. 


REASONS 

1.  Every  Z  has  but  one  bi- 

sector.    §  53. 

2.  By  hyp. 

3.  By  iden. 

4.  By  cons. 

5.  Two  A  are  equal  if   two 

sides  and  the  included 
Z  of  one  are  equal  re- 
spectively to  two  sides 
and  the  included  Z  of 
the  other.  §  107. 

6.  Homol.  parts  of  equal  fig- 

ures are  equal.     §  110. 


112.  Cor.  I.     The  bisector  of  the  angle  at  the  vertex  of 
an  isosceles  triangle  is  perpendicular  to  the  base  and 
bisects  it. 

113.  Cor.   n.     An   equilateral    ftiangle   is   also   equi- 
angular.   

Ex.  68.     The  bisectors  of  the  base  angles  of  an  isosceles  triangle  are 
equal. 


BOOK   I 


31 


114.  Historical  Note.  Exercise  69  is  known  as  the  pons  asi- 
norum,  or  bridge  of  asses,  since  it  has  proved  difficult  to  many  beginners 
in  geometry.  This  proposition 
and  the  proof  here  suggested 
are  due  to  Euclid,  a  great 
mathematician  who  wrote  the 
first  systematic  text-book  on 
geometry.  In  this  work,  known 
as  Euclid's  Elements,  the  exer- 
cise here  given  is  the  fifth  prop- 
osition in  Book  I. 

Of  the  life  of  Euclid  there  is 
but  little  known  except  that  he 
was  gentle  and  modest  and 
"  was  a  Greek  who  lived  and 
taught  in  Alexandria  about  300 
B.C."  To  him  is  attributed  the 
saying,  "  There  is  no  royal  road 
to  geometry. "  His  appreciation 
of  the  culture  value  of  geometry  is  shown  in  a  story  related  by  Stobaeus 
(which  is  probably  authentic).  "A  lad  who  had  just  begun  geometry 
asked,  *  What  do  I  gain  by  learning  all  this  stuff  ?  '  Euclid  called  his 
slave  and  said,  '  Give  this  boy  some  coppers,  since  he  must  make  a 
profit  out  of  what  he  learns.'  " 


EUCLID 


Ex.  69.  By  using  the  accompanying  diagram  prove 
that  the  base  angles  of  an  isosceles  triangle  are  equal. 

HINT.     Prove  A  ABE  =  A  DEC.     Then  prove 
&ACE  =  &DAC. 

Ex.  70.  (a)  If  equal  segments  measured  from  the 
vertex  are  laid  off  on  the  arms  of  an  isosceles  triangle, 
the  lines  drawn  from  the  ends  of  the  segments  to  the  foot 
of  the  bisector  of  the  vertex  angle  will  be  equal. 

(6)  Extend  (a)  to  the  case  in  which  the  equal  segments  are  laid  off  on 
the  arms  prolonged  through  the  vertex. 

Ex.  71.  (a)  If  equal  segjpients  measured  from  the  ends  of  the  base  are 
laid  off  on  the  arms  of  an  isosceles  triangle,  the  lines  drawn  from  the  ends 
of  the  segments  to  the  foot  of  the  bisector  of  the  vertex  angle  will  be  equal. 

(6)  Extend  (a)  to  the  case  in  which  the  equal  segments  are  laid  off  on 
the  arms  prolonged  through  the  ends  of  the  base. 


32 


PLANE   GEOMETRY 


Ex.  72.     (a)  If  equal  segments  measured  from  the  ends  of  the 
are  laid  off  on  the  base  of  an  isosceles  triangle,  the  lines  joining  the  vertex 
of  the  triangle  to  the  ends  of  the  segments  will  be  equal. 

(6)  Extend  (a)  to  the  case  in  which  the  equal  segments  are  laid  off  on 
the  base  prolonged  (Fig.  1) . 


FIG.  1. 


FIG.  2. 


Ex.  73.  (a)  If  equal  segments  measured  from  the  ends  of  the  base 
are  laid  off  on  the  arms  of  an  isoscles  triangle,  the  lines  drawn  from  the 
ends  of  the  segments  to  the  opposite  ends  of  the  base  will  be  equal. 

(6)  Extend  (a)  to  the  case  in  which  the  equal 
segments  are  laid  off  on  the  arms  prolonged 
through  the  ends  of  the  base  (Fig.  2). 

Ex.  74.  Triangle  ABC  is  equilateral,  and 
AE  =  BF  =  CD.  Prove  triangle  EFD  equi- 
lateral . 


A 


C 


115.  Measurement  of  Distances  by  Means 
of  Triangles.  The  theorems  which  prove  triangles  equal  are 
applied  practically  in  measuring  distances  on  the  surface  of 
the  earth.  Thus,  if  it  is  desired  to  find  the  distance  between 
two  places,  A  and  B,  which  are  separated  by  a  pond  or  other 
obstruction,  place  a  stake  at  some  point 
accessible  to  both  A  and  B,  as  F. 
Measure  the  distances  FA  and  FB ;  then, 
keeping  in  line  with  F  and  B,  measure 
CF  equal  to  FB,  find,  in  line  with  F 
and  A,  measure  FE.  equal  to  FA.  Lastly 
measure  CE,  and  the  distance  from  A  to  B  is  thus  obtained,  since 
AB  is  equal  to  CE.  Can  this  method  be  used  when  A  and  B  are 
on  opposite  sides  of  a  hill  and  each  is  invisible  from  the  other  ? 


BOOK  I 


33 


Ex.  75.  Show  how  to  find  the  distance  across  a  river  by  taking  the 
following  measurements.  Measure  a  convenient  dis- 
tance along  the  bank,  as  J?T,  and  fix  a  stake  at  its 
mid-point,  F.  Proceed  at  right  angles  to  BT  from 
T  to  the  point  P,  where  F,  S,  and  P  are  in  line ; 
measure  FT. 

Ex.  76.     An    army    engineer    wished    to    obtain 
quickly  the  approximate  distance  across  a  river,  and 
had  no  instruments  with  which  to  make  measure- 
ments.    He  stood  on  the  bank  of  the  river,   as  at  A,  and  sighted  the 
opposite  bank,   or  B.      Then  without 
raising  or  lowering  his  eyes,  he  faced 


about,  and  his  line  of  sight  struck  the 
ground  at  C.  He  paced  the  distance, 
AC,  and  gave  this  as  the  distance  across  the  river.  Explain  his  method. 


Ex.  77.  Tell  what  measurements  to  make 
to  obtain  the  distance  between  two  inaccessible 
points,  R  and  S  (Fig.  1). 

Ex.  78.  The  fact  that  a  triangle  is  deter- 
mined if  its  base  and  its  base  angles  are  given 
was  used  as  early  as  the  time  of  Thales  (640 
B.C.)  to  find  the  distance  of  a  ship  at  sea;  the 
base  of  the  triangle  was  usually  a  lighthouse 
tower  and  the  base  angles  were  found  by  ob- 
servation. Draw  a  figure  and  explain. 


FIG.  1. 


Ex.   79.     Explain    the   following    method    of  finding    Sfi    (Fig.  2). 
Place  a  stake  at  $,  and  another  at  a  convenient      p 
point,   Q,  in  line  with   S  and  E.     From  a  con- 
venient point,  as  T,  measure  TS  and  TQ.     Pro- 
long QT,  and  make   TF  equal  to  QT.     Prolong 
ST,   and   make    TB  equal   to    ST.     Then   keep 
in  line  with  F  and  B,  until  a  point  is  reached,  as 
G,  where   T  and  ft  come  into  line.     Then   BG 
is  equal  to  the  required  distance,  US. 

Ex.  80.  In  an  equilateral  triangle,  if  two 
lines  are  drawn  from  the  ends  of  the  base, 
making  equal  angles  with  the  base,  the  lines  are 
equal.  Is  this  true  of  every  isosceles  triangle?  FIG.  2. 


34 


PLANE   GEOMETRY 


PROPOSITION  V.    'THEOREM 

116.   Two  triangles  are  equal  if  the  three  sides  of  one 
are  equal  respectively  to  the  three  sides  of  the  other. 


Given  A  ABC  and  RST,  AB  =  RS,  BC=ST,  and  CA  =  TR. 
To  prove  A  ABC  =  A  RST. 


2. 


ARGUMENT 

Place  A  RST  so  that  the 
longest  side  RT  shall  fall 
upon  its  equal  AC,  R 
upon  A,  T  upon  (7,  and 
so  that  S  shall  fall  oppo- 
site B. 

D.raw  BS. 


3.   A  AB sis  isosceles. 

5.    A  BCS  is  isosceles. 
7.   Zl- 


REASONS 

1.  Any  geometric  figure  may 

be  moved  from  one  posi- 
tion to  another  without 
change  of  size  or  shape. 
§  54,  14. 

2.  A  str.  line  may  be  drawn 

from  any  one  point  to 
any  other.  §  54,  15. 

3.  AB  =  RS,  by  hyp. 

4.  The  base  A  of  an  isosceles 

A  are  equal.     §  111. 

5.  BC=ST,  by  hyp. 

6.  Same  reason  as  4. 

7.  If    equals    are    added    to 

equals,  the  sums  are 
equal.  §  54,  2. 


BOOK  I 


35 


ARGUMENT 
8.    .'.  Z  A£C=Z.  CSA. 


9.    .'.  A  ABC=  A 

i.e. 


Q.E.D. 


REASONS 

The  whole  =  the  sum  of 
all  its  parts.  §  54,  11. 

Two  A  are  equal  if  two 
sides  and  the  included  /- 
of  one  are  equal  respec- 
tively to  two  sides  and 
the  included  Z  of  the 
other.  §  107. 


Ex.  81.     («)  Prove  Prop.  V,  using  two  obtuse  triangles  and  applying 
the  shortest  side  of  one  to  the  shortest  side  of  the  other. 

(6)  Prove  Prop.  V,  using  two  right  triangles  and  applying  the  shortest 
side  of  one  to  the  shortest  side  of  the  other. 


117.   Question.     Why  is  not  Prop.  V  proved  by  superposition? 


SUMMARY    OF    CONDITIONS    FOB    EQUALITY    OF    TRIANGLES 


118.   Two  triangles  are  equal  if 


of  one  are  equal  respectively  to 


of  the  other. 


a  side  and  the  two  adjacent 

angles 
two  sides  arid  the  included 

angle 

three  sides 
a  side  and  the  two  adjacent 

angles 
two  sides  and  the  included 

angle 
three  sides 


Ex.  82.  The  median  to  the  base  of  an  isosceles  triangle  bisects  the 
angle  at  the  vertex  and  is  perpendicular  to  the  base. 

Ex.  83.  In  a  certain  quadrilateral  two  adjacent  sides  are  equal ;  the 
other  two  sides  are  also  equal.  Find  a  pair  of  triangles  which  you  can 
prove  equal. 


36  PLANE   GEOMETRY 

Ex.  84.  If  the  opposite  sides  of  a  quadrilateral  are  equal,  the  opposite 
angles  also  are  equal. 

Ex.  85.  If  two  isosceles  triangles  have  the  same  base,  the  line  joining 
their  vertices  bisects  each  vertex  angle  and  is  perpendicular  to  the  com- 
mon base.  (Two  cases.) 

Ex.  86.     In  what  triangles  are  the  three  medians  equal  ? 

Ex.  87.     In  what  triangles  are  two  medians  equal  ? 

Ex.  88.  If  three  rods  of  different  lengths  are  put  together  to  form  a 
triangle,  can  a  different  triangle  be  formed  by  arranging  the  rods  in  a 
different  order  ?  Will  the  angles  opposite  the  same  rods  always  be  the 
same? 

Ex.  89.  If  two  sides  of  one  triangle  are  equal  respectively  to  two 
sides  of  another,  and  the  median  drawn  to  one  of  these  sides  in  the  first 
is  equal  to  the  median  drawn  to  the  corresponding  side  in  the  second,  the 
triangles  are  equal. 

119.  Def.     A  circle  is  a  plane  closed  figure  whose  boundary 
is  a  curve  such  that  all  straight  lines  to  it  from  a  fixed  point 
within  are  equal. 

The  curve  which  forms  the  boundary  of  a  circle  is  called  the 
circumference.  The  fixed  point  within  is  called  the  center,  and 
a  line  joining  the  center  to  any  point  on  the  circumference  is  a 
radius. 

120.  It  follows  from  the  definition  of  a  circle  that : 
All  radii  of  the  same  circle  are  equal. 

121.  Def.     Any  portion  of  a  circumference  is  called  an  arc. 

122.  Assumption  18.     Circle  postulate.      A  circle  may  be  con- 
structed having  any  point  as  center,  and  having  a  radius  equal  to 
any  finite  line. 

123.  The  solution  of  a  problem  of  construction  consists  of 
three  distinct  steps : 

(1)  The  construction,  i.e.  the  process  of  drawing  the  required 
figure  with  ruler  and  compasses. 

(2)  TJie  proof,  a  demonstration  that  the  figure  constructed 
fulfills  the  given  conditions. 

(3)  TJie  discussion,  i.e.  a  statement  of  the  conditions  under 
which  there  may  be  no  solution,  one  solution,  or  more  than  one. 


BOOK  I 


37 


PROPOSITION  VI.     PROBLEM 

124.   To  construct  an  equilateral  triangle,  with  a  given 
line  as  side. 


\D 


Given   line  EC. 

To  construct   an  equilateral  triangle  on  EG. 

I.    Construction 

1.  With  E  as  center  and  EC  as  radius,  construct  circle  CAD. 

2.  With  C  as  center  and  EC  as  radius,  construct  circle  EM  A. 

3.  Connect  point  J,-at  which  the  circumferences  intersect, 
with  E  and  C. 

4.  A  ABC  is  the  required  triangle. 


II.    Proof 


ARGUMENT 


3. 


/.  AB=BC  =  CA. 


A  ABC  is  equilateral. 

Q.E.D. 


REASONS 

1.  All  radii  of  the  same  circle 

are  equal.     §  120. 

2.  Things  equal  to  the  same 

thing  are  equal  to  each 
other.     §  54,  1. 
3    A  A  having  its  three  sides 
equal  is  equilateral.  §  95. 


III.    Discussion 

This  construction  is  always  possible,  and  there  is  only  one 
solution,    (See  §  116.) 


38 


PLANE   GEOMETRY 


PROPOSITION  VII.     PROBLEM 

125.    With  a  given  vertex  and  a  given  side,  to  construct 
an  angle  equal  to  a  given  angle. 


Given   vertex  A,  side  AB,  and  Z  CDE. 

To  construct  an  Z  equal  to  Z  CDE  and  having  A  as  vertex 

and  AB  as  side. 

/ 
I.    Construction 

1.  With  D  as  center,  and  with  any  convenient  radius,  de- 
scribe an  arc  intersecting  the  sides  of  Z  D  at  F  and  G.  respec- 
tively. 

2.  With  A  as  center,  and  with  the  same  radius,  describe  the 
indefinite  arc  IH,  cutting  AB  at  /. 

3.  With  /  as  center,  and  with  a  radius  equal  to  str.  line  FG, 
describe  an  arc  intersecting  the  arc  IH  at  K. 

4.  Draw  AK. 

5.  Z  BAK  =  Z  CDE,  and  is  the  Z  required. 


II.    Proof 


ARGUMENT 
1.   Draw  FG  and  IK. 


2.  In  A  FDG  and  I  A  K,  DF  =  AL 

3.  DG  =  AK. 

4.  FG  =  IK. 


REASONS 

1.  A  str.  line  may  be  drawn 

from   any  one   point  to 
any  other.     §  54,  15. 

2.  By  cons. 

3.  By  cons. 

4.  By  cons. 


BOOK   I 


39 


ARGUMENT 


6.    .'.  /.EAK  — 


Q.E.D. 


REASONS 

5.  Two  A   are   equal   if   the 

three  sides  of  one  are 
equal  respectively  to  the 
three  sides  of  the  other. 
§  116. 

6.  Homol.     parts     of     equal 

figures  are  equal.    §  110. 


III.    Discussion 

This  construction  is  always  possible,  and  there  is  only  one 
solution.  

Ex.  90.     Construct  a  triangle,  given  two  sides  and  the  included  angle. 

Ex.  91.     Construct  an  isosceles  triangle,  given  the  vertex  angle  and 
an  arm. 

-Ex.  92.     Construct  a  triangle,  given  a  side   and  the  two   adjacent 
angles. 

Ex.  93.     Construct  an  isosceles  triangle,  given   an  arm  and  one  of 
the  equal  angles. 

Ex.  94.     How  many  parts  determine  a  triangle  ?     Do  three  angles 
determine  it  ?     Explain. 

Ex.  95.     Construct  an  isosceles  triangle,  given  the  base  and  an  arm. 
"  Ex.  96.     Construct  a  scalene  triangle,  given  the  three  sides. 


126.    Def.     The  bisector  of  an  angle  of  a  triangle  is  the  line 
from  the  vertex  of  the  angle  bisecting  the  angle  and  limited 
by    the    opposite    side    of    the 
triangle.  A 

Ex.  97.  In  what  triangles  are 
the  three  bisectors  equal  ? 

Ex.  98.  In  what  triangles  are 
two  bisectors,  and  only  two,  equal  ? 

Ex.  99.  In  what  triangles  are 
the  medians,  the  bisectors,  and  the 
altitudes  identical  ? 


B 


40 


PLANE   GEOMETRY 


PROPOSITION  VIII.     PROBLEM 
127.   To  construct  the  bisector  of  a  given  angle. 

JC 


V 


Given    Z  AB  C. 

To  construct   the  bisector  of  Z  ABC. 

I.    Construction 

1.  With    B    as    center,    and    with    any    convenient   radius, 
describe  an  arc  intersecting  BA  at  E  and  BC  at  D. 

2.  With  D  and  E  as  centers,  and  with  equal  radii,  describe 
arcs  intersecting  at  F. 

3.  Draw  BF. 

4.  BF  is  the  bisector  of  Z.  ABC. 


II.    Proof 


ARGUMENT 

1.  In   A  EBF  and 

5^  =  BD. 

2.  ^  =  Z)*7. 

3.  BF  =  BF. 

4.  .-.  AEBF=  AFBD 


5.    .'. /.  EBF  =  Z.FBD. 


REASONS 


1.  By  cons. 

2.  By  cons. 

3.  By  iden. 

4.  Two  A  are   equal    if   the 

three  sides  of  one  are 
equal  respectively  to  the 
three  sides  of  the  other. 
§116. 

5.  Homol.     parts     of     equal 

figures  are  equal.    §  110. 


BOOK   I 


41 


ARGUMENT 

6.    .'.  BF    is    the   bisector   of 
Z  ABC. 

Q.E.D. 


REASONS 

6.  The  bisector  of  an  Z  is 
the  line  which  divides 
the  Z  into  two  equal 
Zs.  §  53. 


III.    Discussion 

This  construction  is  always  possible,  and  there  is  only  one 
solution. 

"  Ex.  100.  Draw  an  obtuse  angle  and  divide  it  into :  (a)  four 
equal  angles  ;  (&)  eight  equal  angles. 

Ex.  101.  Construct  the  bisector  of  the  vertex  angle  of  an  isosceles 
triangle. 

Ex.  102.  Draw  two  intersecting  lines  and  construct  the  bisectors  of 
the  four  angles  formed. 

Ex.  103.  Bisect  an  angle  between  two  bisectors  in  Ex.  102,  and  find 
the  number  of  degrees  in  each  angle. 

Ex.  104.  Construct  the  bisector  of  an  ex^rior  angle  at  the  base  of  an 
isosceles  triangle. 

Ex.  105.  Construct  the  bisectors  of  the  three  angles  of  any  triangle. 
What  can  you  infer  about  them  ?  Can  the  correctness  of  this  inference 
be  proved  by  making  a  careful  construction  ? 


128.  Def.  The  distance  between  two  points 
is  the  length  of  the  straight  line  joining  them. 
Thus  if  three  points,  A,  B,  and  (7,  are  so  located 
that  AB  =  AG,  A  is  said  to  be  equidistant  from 
B  and  G. 


Ex.  106.     Find  all   the  points  on    the   blackboard 
which  are  one  foot  from  a  h'xed  point,  P,  on  the  blackboard. 

Ex.  107.  Draw  a  line,  AB,  on  the  blackboard  and  mark  some  point 
near  the  line,  as  P.  Find  all  the  points  in  AB  that  are  a  foot  from  P. 

Ex.  108.  Mark  a  point,  Q,  on  the  blackboard.  Find  all  the  points  on 
the  blackboard  which  are :  (a)  ten  inches  from  Q  ;  (6)  four  inches  from 
Q.  How  far  are  the  points  of  (6)  from  the  points  of  (a),  if  the  distance 
is  measured  on  a  line  through  Q  ? 


42  PLANE   GEOMETRY 

LOCI 

129.  In  many  geometric  problems  it  is  necessary  to  locate 
all  points  which  satisfy  certain  prescribed   conditions,  or  to 
determine  the  path  traced  by  a  point  which  moves  according 
to  certain  fixed  laws.     Thus,  the  points  in  a  plane  two  inches 
from  a  given  point  are  in  the  circumference  of  a  circle  whose 
center  is  the  given  point  and  whose  radius  is  two  inches. 

Again,  let  it  be  required  to  find  all  points  in  a  plane  two 
inches  from  one  fixed  point  and  three  inches  from  another. 
All  points  two  inches  from  the  fixed  point  P  are  in  the  cir- 
cumference of  the  circle  LMS,  having  P  for  center  and  having 
a  radius  equal  to  two 
inches.  All  points  three 
inches  from  the  fixed 
point  Q  are  in  the  cir- 
cumference of  the  circle 
LRT,  having  Q  for  center 
and  having  a  radius 
equal  to  three  inches. 
If  the  two  circles  are 
wholly  outside  of  each  other;  there  will  be  no  points  satisfying 
the  two  prescribed  conditions ;  if  the  two  circumferences  touch, 
but  do  not  intersect,  there  will  be  one  point;  if  the  two  circum- 
ferences intersect,  there  will  be  two  points.  It  will  be  proved 
later  (§  324)  that  there  cannot  be  more  than  two  points  which 
satisfy  both  of  the  given  conditions. 

130.  Def.     A  figure  is  the  locus  of  all  points  which  satisfy 
one  or  more  given  conditions,  if  all  points  in  the  figure  satisfy 
the  given  conditions  and  if  these  conditions  are  satisfied  by  no 
other  points. 

A  locus,  then,  is  an  assemblage  of  points  which  obey  one  or 
more  definite  laws. 

It  is  often  convenient  to  locate  these  points  by  thinking  of 
them  as  the  path  traced  by  a  moving  point  the  motion  of  which 
is  controlled  by  certain  fixed  laws. 


BOOK   I  43 

131.  In  plane  geometry  a  locus  may  be  composed  of  one  or 
more  points  or  of  one  or  more  lines,  or  of  any  combination  of 
points  and  lines. 

132.  Questions.*  —  What  is  the  locus  of  all  points  in  space  two 
inches  from  a  given  point  ?     What  is  the  locus  of  all  points  in  space  two 
inches  from  a  given  plane  ?     What  is  the  locus  of  all  points  in  space  such 
that  perpendiculars  from  them  to  a  given  plane  shall  be  equal  to  a  given 
line  ?     What  is  the  locus  of  all  points  on  the  surface  of  the  earth  midway 
between  the  north  and  south  poles  ?  23£°  from  the  equator  ?  23£°  from 
the  north   pole  ?   90°  from  the   equator  ?     What  is  the  locus  of  a  gas 
jet  four  feet  from  the  ceiling  of  this  room  ?    four  feet  from  the  ceiling 
and  five  feet  from  a  side  wall  ?    four  feet  from  the  ceiling,  five  feet 
from  a  side  wall,  and  six  feet  from  an  end  wall  ? 


Ex.  109.  Given  an  unlimited  line  AB  and  a  point  P.  Find  all  points 
in  AB  which  are  also  :  (a)  three  inches  from  P;  (6)  at  a  given  distance, 
a,  from  P. 

Ex.  110.  Given  a  circle  with  center  O  and  radius  six  inches.  State, 
without  proof,  the  locus  :  (a)  of  all  points  four  inches  from  0 ;  (&)  of 
all  points  five  inches  from  the  circumference  of  the  circle,  measured  on 
the  radius  or  radius  prolonged. 

Ex.  111.  Given  the  base  and  one  adjacent  angle  of  a  triangle,  what  is 
the  locus  of  the  vertex  of  the  angle  opposite  the  base  ?  (State  without 
proof.) 

Ex.  112.  Given  the  base  and  one  other  side  of  a  triangle,  what  is  the 
locus  of  the  vertex  of  the  angle  opposite  the  base  ?  (State  without  proof.) 

Ex.  113.  Given  the  base  and  the  other  two  sides  of  a  triangle,  what 
is  the  locus  of  the  vertex  of  the  angle  opposite  the  base  ? 

Ex.  114.  Given  the  base  of  a  triangle  and  the  median  to  the  base,  what 
is  the  locus  of  the  end  of  the  median  which  is  remote  from  the  base  ? 

Ex.  115.  Given  the  base  of  a  triangle,  one  other  side,  and  the 
median  to  the  base,  what  is  the  locus  of  the  vertex  of  the  angle  opposite 
the  base  ? 

133.  Question.  In  which  of  the  exercises  above  was  a  triangle 
determined  9 

*  In  order  to  develop  the  imagination  of  the  student  the  authors  deem  it  advisable 
in  this  article  to  introduce  questions  involving  loci  in  space.  It  should  be  noted  that 
no  proofs  of  answers  to  these  questions  are  demanded.  V 


44 


PLANE   GEOMETRY 


PROPOSITION  IX.     THEOREM 

134.   Every  point   in  the    perpendicular   bisector  of  a 
line  is  equidistant  from  the  ends  of  that  line. 


A  C  B 

Given   line  AB,  its  _L  bisector  CD,  and  P  any  point  in  CD. 
To  prove   PA  =  PB. 

ARGUMENT  REASONS 


1.    In  A  APC  and  CPB, 


PC=PC. 
Z  PGA  =  Z  BCP. 
A  APC  =  A  CPB. 


PA  = 


Q.E.  D. 


1.  By  hyp. 

2.  By  iden. 

3.  All  rt.  A  are  equal.     §  64. 

4.  Two  A  are  equal  if   two 

sides  and  the  included 
Z  of  one  are  equal  re- 
spectively to  two  sides 
and  the  included  Z  of 
the  other.  §  107. 

5.  Hoinol.  parts  of  equal  fig- 

ures are  equal.     §  110. 


Ex.  116.  Four  villages  are  so  located  that  B  is  25  miles  east  of  A, 
C  20  miles  north  of  A,  and  D  20  miles  south  of  A.  Prove  that  B  is  as 
far  from  C  as  it  is  from  D. 

Ex.  117.  In  a  given  circumference,  find  the  points  equidistant  from 
two  given  points,  4  and  JS, 


BOOK  I  45 

135.  Def.     One  theorem  is  the  converse  of  another  when  the 
conclusion  of  the  first  is  the  hypothesis  of  the  second,  and  the 
hypothesis  of  the  first  is  the  conclusion  of  the  second. 

The  converse  of  a  truth  is  not  always  true  ;  thus,  "  All  men 
are  bipeds  "  is  true,  but  the  converse,  "  All  bipeds.are  men,"  is 
false.  "All  right  angles  are  equal"  is  true,  but  "All  equal 
angles  are  right  angles"  is  false. 

136.  Def.     One  theorem  is  the  opposite  of  another  when  the 
hypothesis  of  the  first  is  the  contradiction  of  the  hypothesis 
of  the  second,  and  the  conclusion  of  the  first  is  the  contradic- 
tion of  the  conclusion  of  the  second. 

The  opposite  of  a  truth  is  not  always  true ;  thus,  "  If  a 
man  lives  in  the  city  of  New  York,  he  lives  in  New  York  State" 
is  true,  but  the  opposite,  "  If  a  man  does  not  live  in  the  city  of 
New  York,  he  does  not  live  in  New  York  State,"  is  false. 

137.  Note.     If  the  converse  of  a  proposition  is  true,  the  opposite  also 
is  true  ;  so,  too,  if  the  opposite  of  a  proposition  is  true,  the  converse  also 
is  true. 

This  may  be  evident  to  the  student  after  a  consideration  of  the  fol- 
lowing type  forms : 

(1)  DIRECT                             (2)  CONVERSE  (3)  OPPOSITE 

If         A  is  B,                 If         C  is  D,  If         A  is  not  B, 

Then    C  is  D.                 Then  A  is  B.  Then   C  is  not  D. 

If  (2)  is  true,  then  (3)  must  be  true.  Again,  if  (3)  is  true,  then  (2) 
must  be  true. 

138.  A  necessary  and  sufficient  test  of  the  completeness  of  a 
definition  is  that  its  converse  shall  also  be  true.     Hence  a 
definition  may  be  quoted  as  the  reason  for  a  converse  or  for  an 
opposite  as  well  as  for  a  direct  statement  in  an  argument. 


Ex.  118.  State  the  converse  of  the  definition  for  equal  figures ; 
straight  line  ;  plane  surface. 

Ex.  119.  State  the  converse  of:  If  one  straight  line  meets  another 
straight  line,  the  sum  of  the  two  adjacent  angles  is  two  right  angles. 

Ex.  120.     State  the  converse  and  opposite  of  Prop.  IX. 

Ex.  121.     State  the  converse  of  Prop.  I.     Is  it  true? 


46 


PLANE   GEOMETRY 
PROPOSITION  X.     THEOREM 


(Converse  of  Prop.  IX) 

139.   Every  point  equidistant  from  the  ends  of  a  line 
lies  in  the  perpendicular  bisector  of  that  line. 


Given   line  US,  and  point  Q  such  that  QR  =  QS. 
To  prove   that  Q  lies  in  the  _L  bisector  of  RS. 


ARGUMENT 

1.  Let  QT  bisect  Z  RQS. 

2.  QR  =  QS. 

3.  .'.A  RQS  is  isosceles. 

4.  .'.  QT  is  the  _L  bisector  of 

RS. 


5.   .'.  Q  lies  in  the  _1_  bisector 

Of  RS.  Q.E.D. 


REASONS 

1.  Every  Z.  has  but   one  bi- 

sector.    §  53. 

2.  By  hyp. 

3.  A  A  having  two  sides  equal 

is  an  isosceles  A.    §  94. 

4.  The  bisector  of   the  Z  at 

the  vertex  of  an  isosceles 
A  is  ±  to  the  base  and 
bisects  it.  §  112. 

5.  By  proof. 


140.  Cor.  I.  Every  point  not  in  the  perpendicular  bi- 
sector of  a  line  is  not  equidistant  from  the  ends  of  the 
line. 

HINT.  Use  §  137,  or  contradict  the  conclusion  and  tell  why  the  con- 
tradiction is  false. 


BOOK  I  47 

141.  Cor.  II.     The  locus  of  all  points  equidistant  from 
the  ends  of  a  given  line  is  the  perpendicular  bisector  of 
that  line. 

HINT.     See  §§  143  and  144. 

142.  Cor.  III.    Two  points  each  equidistant  from  the  ends 
of  a  line  determine  the  perpendicular  bisector  of  tlie  line. 

HINT.     Use  §  139  and  §  25. 

143.  In  order  to  prove  that  a  locus  problem  is  solved  it  is 
necessary  and  sufficient  to  show  two  things: 

(1)  That  every  point  in   the   proposed   locus   satisfies   the 
prescribed  conditions. 

(2)  That  every  point  outside  of  the  proposed  locus  does  not 
satisfy  the  prescribed  conditions. 

Instead  of  proving  (2),  it  may  frequently  be  more  convenient 
to  prove : 

(2')  That  «very  point  which  satisfies  the  prescribed  con- 
ditions lies  in  the  proposed  locus. 

144.  Note.     In  exercises  in  which  the  student  is  asked  to  "Find 
a  locus,"  it  must  be  understood  that  he  has  not  found  a  locus  until  he 
has  given  a  proof  with  regard  to  it  as  outlined  above.     The  proof  must  be 
based  upon  a  direct  proposition  and  its  opposite  ;  or,  upon  a  direct  propo- 
sition and  its  converse, 

Ex.  122.  Find  the  locus  of  all  points  equidistant  from  two  given 
points  A  and  B. 

.         Ex.  123.     In  a  given  unlimited  line  AB,  find  a  point  equidistant  from 
two  given  points  C  and  D  not  on  this  line. 

Ex.  124.  Given  a  circle  with  center  O,  also  a  point  P.  Find  all 
points  which  lie  in  the  circumference  of  circle  0,  and  which  are  also  (a) 
two  inches  from  P ;  (6)  a  distance  of  d  from  P. 

Ex.  125.  Find  all  points  at  a  distance  of  d  from  a  given  point  P,  and 
at  the  same  time  at  a  distance  of  m  from  a  given  point  Q. 

Ex.  126.     Given  a  circle  0  with  radius  r.    Find  the  locus  of  the  mid- 
^  points  of  the  radii  of  the  circle. 

Ex.  127.  Given  two  circles  having  the  same  center.  State,  without 
proof,  the  locus  of  a  point  equidistant  from  their  circumferences. 

JEx.  128.  The  perpendicular  bisector  of  the  base  of  an  isosceles  tri- 
mgle  passes  through  the  vertex. 


48 


PLANE   GEOMETRY 


PROPOSITION  XL     PROBLEM 

145.   To  construct  the  perpendicular  bisector  of  a  given 
straight  line. 


£\ 

/* 

\ 

/ 

\ 

1 

\ 

A 

\                 B  A 

B 

i 

K\ 

>                     \ 

(C 

\ 

i                                              t 

\ 

/ 

\                                             \ 

/ 

\ 

V 

Fi 

a.  1.                                             FIG 

.  2. 

Given   line  AB  (Fig.  1). 

To  construct   the  perpendicular  bisector  of  AB. 


I.    Construction 

1.  With  A  as  center,  and  with  a  convenient  radius  greater 
than  half  AB,  describe  the  arc  EF. 

2.  With  B  as  center,  and  with  the  same  radius,  describe  the 
arc  HK. 

3.  Let  C  and  D  be  the  points  of  intersection  of  these  two 
arcs. 

4.  Connect  points  C  and  D. 

5.  CD  is  the  J_  bisector  of  AB. 

II.    The  proof  and  discussion  are  left  as  an  exercise  for  the 
student. 

HINT.     Apply  §  142. 

146.  Note.     The  construction  given  in  Fig.  2  may  be  used  when  the 
position  of  the  given  line  makes  it  more  convenient. 

147.  Question.     Is  it  necessary  that  GA  shall  equal   CB?  that  CA 
shall  equal  DA  (Fig.  1)  ?     Give  the  equations  that  must  hold. 


BOOK  I  40 


PROPOSITION  XII.     PROBLEM 

148.   To  construct  a  perpendicular  to  a  given  straight 
line  at  a  given  point  in  the  line. 


> 

< 

1 

\ 

A         D\        C        IE 

B 

Given   line  AB  and  point  C  in  the  line. 
To  construct    a  _L  to  AB  at  C- 

I.    Construction 

1.  With  C  as  center,  and  with  any  convenient  radius,  draw 
arcs  cutting  AB  on  each  side  of  C7,  as  at  D  and  E. 

2.  Then  with  D  and  E  as  centers,  and  with  a  longer  radius, 
draw  two  arcs  intersecting  each  other  at  F. 

3.  Draw  Fa 

4.  FC  is  _L  to  AB  at  C. 

II.    The  proof  and  discussion  are  left  as  an  exercise  for  the 
student. 

Ex.  129.     Construct  the  perpendicular  bisector  of  a  line  given  at  the 
bottom  of  a  page  or  of  a  blackboard. 

Ex.   130.     Divide  a  given  line  into  four  equal  parts. 

Ex.   131.     Construct  the  three  medians  of  a  triangle. 

Ex.   132.     Construct  a  perpendicular  to  a  line  at  a  given  point  when 
the  given  point  is  one  end  of  the  line.     (HINT.     Prolong  the  line.) 

Ex.  133.     Construct  a  right  triangle,  given  the  two  arms  a  and  />. 
^*Ex.  134.     Construct  a  right  triangle,  given  the  hypotenuse  and  an  arm. 

Ex.  135.     Construct  the  complement  of  a  given  angle. 

Ex.  136.     Construct  an  angle  of  45°;  of  135°. 

Ex.  137.     Construct  a  quadrilateral,  given  four  sides  and  the  angle 
,  between  two  of  them. 


PLANE   GEOMETRY 


PROPOSITION  XIII.     PROBLEM 

149.   From,  a  point  outside  a  line  to  construct   a  per- 
pendicular  to  the  line. 


\C 


di 


Given   line  AB  and  point  P  outside  of  AB. 
To  construct    a  _L  from  P  to  AB. 

I.    Construction 

1.  With  P  as  center,  and  with  a  radius  of  sufficient  length, 
describe  an  arc  cutting  AB  at  points  C  and  D. 

2.  With  C  and  D  as  centers,  and  with  any  convenient  radius, 
describe  arcs  intersecting  at  Q. 

3.  Draw  PQ. 

4.  PQ  is  a  _L  from  P  to  AB. 

II.  The  proof  is  left  as  an  exercise  for  the  student.  The 
discussion  will  be  given  in  §  154. 

150.  Question.  Must  P  and  Q  be  on  opposite  sides  of  AB  ?  Is  it, 
necessary  that  PC=QC?  

Ex.  138.  Construct  the  three  altitudes  of  an  acute  triangle.  Do  they 
seem  to  meet  ?  where  ? 

Ex.  139.  Construct  the  three  altitudes  of  a  right  triangle.  Where  do 
they  seem  to  meet  ? 

Ex.  140.  Construct  the  three  altitudes  of  an  obtuse  triangle.  Where 
do  they  seem  to  meet  ? 

Ex.  141.  Construct  a  triangle  ABO,  given  two  sides  and  the  median 
drawn  to  one  of  them.  Abbreviate  thus :  given  a,  />,  and  ma. 


BOOK  I  51 

151.  Analysis  of  a  problem  of  construction.     In  the  more 
difficult  problems  of  construction  a  course  of  reasoning  is  some- 
times necessary  to  enable  the  student  to  discover  the  process  of 
drawing  the  required  figure.     This  course  of  reasoning  is  called 
the  analysis  of  the  problem.     It  is  illustrated  in  §  152  and  is 
more  fully  treated  in  the  exercises  following  §  274. 

152.  Note.     In  such  problems  as  Ex.  141,  it  is  well  first  to  imagine 
the  problem  solved  and  to  sketch  a  figure  to  represent  the  desired  con- 
struction.    Then  mark  (with  colored  crayon,  if  convenient)  the  parts 
supposed  to  be  given.     By  studying  carefully  the  relation  of  the  given 
parts  to  the  whole  figure^  try  to  find  some  part  of  the  figure  that  you  can 
construct.     This    will    generally   be    a    triangle. 

After  this  part  is  constructed  it  is  usually  an  easy 
matter  to  complete  the  required  figure. 

Thus :  Problem.  Let  it  be  required  to  con- 
struct an  isosceles  triangle,  given  an  arm  and  the 
altitude  upon  it.  By  studying  the  figure  with 
the  given  parts  marked  (heavy  or  with  colored 

crayon),  it  will  be  seen  that  the  solution  of  the        ____^___J 
problem  depends  in  this  case  upon  the  construe-      A  C 

tion  of  a  right  triangle,  given  the  hypotenuse  and 

one  arm.  The  right  triangle  ABD  may  now  be  constructed,  and  it  will 
be  readily  seen  that  to  complete  the  construction  it  is  only  necessary  to 
prolong  BD  to  (7,  making  BC  =  AB,  and  to  connect  A  and  C. 


Ex.  142.  Construct  a  triangle  .ABC,  given  two  sides  and  an  altitude 
to  one  of  the  given  sides.  Abbreviate  thus  :  given  a,  b,  and  hb. 

Ex.  143.  Construct  a  triangle  ABC,  given  a  side,  an  adjacent  angle, 
and  the  altitude  to  the  side  opposite  the  given  angle.  Abbreviate  thus : 
given  a,  B,  and  h(>. 

Ex.  144.  Construct  an  isosceles  triangle,  given  an  arm  and  the  me- 
dian to  it. 

Ex.  145.  Construct  an  isosceles  triangle,  given  an  arm  and  the  angle 
which  the  median  to  it  makes  with  it. 

Ex.  146.  '  Construct  a  triangle,  given  two  sides  and  the  angle  which  a 
median  to  one  side  makes  :  (a)  with  that  side  ;  (b)  with  the  other  side. 

Ex.  147.  Construct  a  triangle,  given  a  side,  an  adjacent  angle,  and 
its  bisector. 


52 


PLANE   GEOMETRY 
PROPOSITION  XIV.     THEOREM 


153.  If  one  side  of  a  triangle  is  prolonged,  the  exterior 
angle  formed  is  greater  than  either  of  the  remote  in- 
terior angles. 

B  F 


\ 


D 


Given    A  ABC  with  AC  prolonged  to  7),  making  exterior  Z  DCB. 
To  prove     Z  DCB  >  Z  ABC  or  Z  CAB. 


ARGUMENT 

1.  Let  E  be  the  mid-point  of 

BC;  draw  AE,  and  pro- 
long it  to  F,  making  EF 
=  AE.  Draw  CF. 

2.  In  A  ^££  and  EFC, 

BE  =  EC. 

3.  AE  =  EF. 

4.  Z  ASM  ='Z 


5.         .' .  A  ABE  =  A  EFC. 


6. 


REASONS 

1.  A  str.  line  may  be  drawn 
from  any  one  point  to 
any  other.  §  54,  15. 


2. 


mid- 


By   cons.  E   is  the 
point  of  BC. 

3.  By  cons. 

4.  If    two    str.   lines   inter- 

sect, the  vertical  A  are 
equal.  §  77. 

5.  Two  A  are  equal  if  two 

sides  and  the  included 
Z  of  one  are  equal  re- 
spectively to  two  sides 
and  the  included  Z  of 
the  other.  §  107. 

6.  Homol.  parts  of  equal  fig- 

ures are  equal.     §  110. 


BOOK  I 


53 


ARGUMENT 
Z  DOB  >  Z  FCE. 


9.   Likewise,    if  BG  is   pro- 
longed to    G, 
/iCAB. 

10.  But  Z  Z><7£  =  Z 

11.  .'.  Z  ZX75  >  Z 


12. 


REASONS 

7.  The  whole  >  any  of  its 

parts.     §  54,  12. 

8.  Substituting  Z  B   fqr   its 

equal  Z  ,F<7J£. 

9.  By  bisecting  line  AC,  and 

by  steps  similar  to  1-8. 


10.  Same  reason  as  4. 

11.  Substituting 

its  equal  Z  ACG. 

12.  By  proof. 


for 


>  /.ABC 
or  Z  CAB.  Q.E.D 

154.  Cor.  From  a  point  outside  a  line 
there  exists  only  one  perpendicular  to 
the  line. 

HINT.    If  there  exists  a  second  JL  to  AB  from   A       c     K  " 

P,  as  PC,  then  Z  PGA  and  /.  PKA  are  both  rt.  A 
and  are  therefore  equal.     But  this  is  impossible  by  §  153 

155.  §§  149  and  154  may  be  combined  in  one  statement  as 
follows : 

From  a  point  outside  a  line  there  exists  one  and  only  one  per- 
pendicular to  the  line. 


Ex.  148.    A  triangle  cannot  contain  two  right  angles. 
Ex.  149.     In  the  figure  of  Prop.  XIV,  is  angle  DGB  necessarily  greater 
than  angle  BCA  ?    than  angle  B  ?  4 

Ex.  150.     In  Fig.  1,  prove  that : 

(1)  Angle  1  is  greater  than  angle   GAE  or 

angle  AEG ; 

(2)  Angle  5  is  greater  than  angle  CBA  or 

angle  BA  E ; 

(3)  Angle  EDA  is  greater  than  angle  3  ; 

(4)  Angle  4  is  greater  than  angle  DAE. 

Ex.  151.  In  Fig.  2,  show  that  angle  6  is 
greater  than  angle  7 ;  also  that  angle  9  is 
greater  than  angle  A,  FIG.  2. 


C  D    E 

FIG.  1. 
A 


D 


B 


54 


PLANE   GEOMETRY 


PROPOSITION  XV.     THEOREM 

156.  If  two  sides  of  a  triangle  are  unequal,  the  anglv 
opposite  tJie  greater  side  is  greater  than  the  angle  opposite 
the  less  side. 


Given    A  ABC  with  BG  >  BA 
To  prove    Z  CAB  >  Z  C. 

ARGUMENT 

1.  On  BC  lay  off  BD  =  AB. 

2.  Draw  AD. 

3.  ThenZl  =  Z2. 

4.  Now  Z2  >  Z<7. 


REASONS 

1.  Circle  post.     §§  122,  157. 

2.  Str.  line  post.  I.     §  54,  15. 

3.  The  base  A  of  an  isosceles 

A  are  equal.     §  111. 

4.  If  one  side  of  a  A  is  pro- 

longed, the  ext.  Z  formed 
>  either  of  the  remote 
int.  Zs.  §  153. 

5.  Substituting     Z  1    for   its 

equal  Z  2. 

6.  The  whole    >    any  of  its 

parts.     §  54,  12. 

7.  If  three  magnitudes  of  the 

same  kind  are  so  related 
that  the  first  >  the  sec- 
ond and  the  second  > 
the  third,  then  the  first 
Q.E.D.  >  the  third.  §  54,  10. 

157.    Note.      Hereafter   the   student  will   not  be  required   to  state 
postulates  and  definitions  in  full  unless  requested  to  do  so  by  the  teacher. 


5.  .-.Zl  >  Z(7. 

6.  But  Z  CAB  >  Z  1. 

7.  /.  Zc^£  >  Z  (7. 


BOOK  I  55 

158.  Note.  When  two  magnitudes  are  given  unequal,  the  laying 
off  of  the  less  upon  the  greater  will  often  serve  as  the  initial  step  in  de- 
veloping a  proof.  

>^  Ex.  152.    Given  the  isosceles  triangle  RST,  with  8T  the  base  and  RT 
^prolonged  any  length,  as  to  K.     Prove  angle  KSll  greater  than  angle  K. 

^   Ex.  153.    If  two  adjacent  sides  of  a  quadrilateral  are  greater  respec- 
^tively  than  the  other  two  sides,  the  angle  included  between   the  two 
shorter  sides  is  greater  than  the  angle  between  the  two  greater  sides. 

^  Ex.  154.     If  from  a  point  within  a  triangle  lines  are  drawn  to  the  ends 
'of  one  of  its  sides,  the  angle  between  these  lines  is  greater  than  the  angle 
between  the  other  two  sides  of  the  triangle.     (See  Ex.  151.) 


159.  The  indirect  method,  or  proof  by  exclusion,  consists  in 
contradicting  the  conclusion  of  a  proposition,  then  showing  the 
contradiction  to  be  false.     The  conclusion  of  the  proposition  is 
thus   established.     This   process   requires   an   examination   of 
every,  possible  contradiction  of  the  conclusion.      For  example, 
to  prove  indirectly  that  A  equals  B  it  would  be  necessary  to 
consider  the  only  three  suppositions  that  are  admissible  in  this 
case,  viz. :  (1)  ^  >  B, 

(2)  A  <  B, 

(3)  A  =  B. 

By  proving  (1)  and  (2)  false,  the  truth  of  (3)  is  established, 
i.e.  A  =  3.  This  method  of  reasoning  is  called  reductio  ad 
absurdum.  It  enables  us  to  establish  a  conclusion  by  showing 
that  every  contradiction  of  it  leads  to  an  absurdity.  Props. 
XVI  and  XArII  will  be  proved  by  the  indirect  method. 

160.  Question.     Would  it  be  possible  to  base  a  proof  upon  a  contra- 
diction of  the  hypothesis  ? 

161.  (a)  In  the  use  of  the  indirect  method  the  student  should  give, 
as  argument  1,  all  the  suppositions  of  which  the  case  he  is  considering 
admits,  including  the  conclusion.     As  reason  1  the  number  of  such  possi- 
ble suppositions  should  be  cited. 

(&)  As  a  reason  for  the  last  step  in  the  argument  he  should  state  which 
of  these  suppositions  have  been  proved  false. 


56 


PLANE   GEOMETRY 


PROPOSITION  XVI.     THEOREM 
(Converse  of  Prop.  IV) 

162.   If  two  angles  of  a  triangle  are  equal,  the  sides 
opposite  are  equal. 


Given    A  RST  with  Z R  =  Z  T. 
To  prove    r  =  t. 

ARGUMENT 
1.    r  >t,   r<t,   or  r  =  t. 


2.   First  suppose    r  >  t ; 
then  /.R>  Z.T. 


3.  This  is  impossible. 

4.  Next  suppose    r  <  t] 

then  Z.R</.T. 

5.  This  is  impossible. 

6.  .'.r  =  t. 


Q.E.D. 


REASONS 

1.  In    this    case    only   thiee 

suppositions  are  admis- 
sible. 

2.  If  two   sides   of  a  A  are 

unequal,  the  Z  opposite 
the  greater  side  >  the 
Z  opposite  the  less  side. 
§  156. 

3.  By  hyp.,  ^R  =  Z.T. 

4.  Same  reason  as  2. 

5.  Same  reason  as  3. 

6.  The  two  suppositions,  r>t 

and  r  <  t,  have  been 
proved  false. 


163.   Cor.  An  equiangular  triangle  is  also  equilateral. 


Ex.  155.  The  bisectors  of  the  base  angles  of  an  isosceles  triangle  form 
an  isosceles  triangle. 


BOOK  I 
PROPOSITION  XVII.     THEOREM 


57 


(Converse  of  Prop.  XV) 

164.  If  two  angles  of  a  triangle  are  unequal,  the  side 
opposite  the  greater  angle  is  greater  than  the  side  opposite 
the  less  angle. 


Given    A  AB C  with  Z  A  >  Z  C. 
To  prove    a  >  c. 

ARGUMENT 

1.  a  <  c,   a  =  c,   or  a  >  c. 

2.  First  suppose  a  <  c ; 

then  Z.A<Z.C. 


3.  This  is  impossible. 

4.  Next  suppose    a  =  c ; 

then  Z4  =  Z.C. 

5.  This  is  impossible. 

6.  .-.  a  >  c. 


Q.E.D. 


REASONS 

1.  In  this  case  only  three  sup- 

positions are  admissible. 

2.  If  two  sides  of  a  A  are  un- 

equal, the  Z.  opposite 
the  greater  side  >  the 
Z  opposite  the  les>-side. 
§156. 

3.  By  hyp.,  /.A  >  Zc. 

4.  The  base  A  of  an  isosceles 

A  are  equal.     §  111. 

5.  Same  reason  as  3. 

6.  The  two  suppositions,  a  <c 

and  a  =  c,  have  been 
proved  false. 


165.  Cor.     The  perpendicular  is  the  shortest  straight 
line  from  a  point  to  a  line. 

166.  Def.     The  length  of  the  perpendicular  from  a  point  to 
a  line  is  called  the  distance  from  the  point  to  the  line. 


58  PLANE   GEOMETRY 

Ex.  156.  The  sum  of  the  altitudes  of  any  triangle  is  less  than  the 
perimeter  of  the  triangle. 

Ex.  157.  Given  the  quadrilateral  ABCD  with  B  and  D  right  angles 
and  BC  greater  than  CD.  Prove  AD  greater  than  AB. 

Ex.158.  Given  triangle  ABC,  with  AC>  BC.  Let  bisectors  of 
angles  A  and  B  meet  at  O.  Prove  AO  >  BO. 

Ex.  159.  A  line  drawn  from  the  vertex  of  an  isosceles  triangle  to  any 
point  in  the  base  is  less  than  one  of  the  equal  sides  of  the  triangle. 

''  Ex.  160.  If  ABC  and  ABD  are  two  triangles  on  the  same  base  and 
on  the  same  side  of  it  such  that  AC  =  BD  and  AD  =  BO,  and  if  AC  and 
BD  intersect  at  0,  prove  triangle  AOB  isosceles. 

Ex.  161.     Prove  Prop.  XVI  by  using  the  figure  and  method  of  Ex.  69. 

'  Ex.  162.  Upon  a  given  base  is  constructed  a  triangle  one  of  the  base 
angles  of  which  is  double  the  other.  The  bisector  of  the  larger  base  angle 
meets  the  opposite  side  at  the  point  P.  Find  the  locus  of  P. 

Ex.  163.  If  the  four  sides  of  a  quadrilateral  are  equal,  its  diagonals 
bisect  each  other. 

Ex.  164.  The  diagonals  of  an  equilateral  quadrilateral  are  perpen- 
dicular to  each  other,  and  they  bisect  the  angles  of  the  quadrilateral. 

Ex.  165.  If  two  adjacent  sides  of  a  quadrilateral  are  equal  and  the 
other  two  sides  are  equal,  one  diagonal  is  the  perpendicular  bisector  of 
the  other.  Tell  which  one  is  the  bisector  and  prove  the  correctness  of 
your  answer. 

Ex.  166.  If,  from  a  point  in  a  perpendicular  to  a  line,  oblique  lines 
are  drawn  cutting  off  equal  segments  from  the  foot  of  the  perpendicular, 
the  oblique  lines  are  equal. 

Ex.  167.     State  and  prove  the  converse  of  Ex.  166. 

Ex.  168.  If,  from  a  point  in  a  perpendicular  to  a  line,  oblique  lines  are 
drawn  cutting  off  unequal  segments  from  the  foot  of  the  perpendicular, 
the  oblique  lines  are  unequal.  Prove  by  laying  off  the  less  segment  upon 
the  greater.  Then  use  Ex.  166,  §  153,  and  §  164. 

Ex.  169.  If,  from  any  point  in  a  perpendicular  to  a  line,  two  unequal 
oblique  lines  are  drawn  to  the  line,  the  oblique  lines  will  cut  off  unequal 
segments  from  the  foot  of  the  perpendicular.  Prove  by  the  indirect  method. 

Ex.  170.  By  means  of  Prop.  XIV,  prove  that  the  sum  of  any' two 
angles  of  a  triangle  is  less  than  two  right  angles. 

Ex.  171.  Construct  a  triangle  ABC,  given  two  sides,  a  and  6,  and 
the  altitude  to  the  third  side,  c.  (See  §  152.) 


BOOK   I 


59 


PROPOSITION  XVIII.     THEOREM 
167.    The  sum  of  any  two  sides  of  a  triangle  is  greater 


than  the  third  side. 


A                        b 

Given  A  ABC. 
To  prove    a  -f  c  >  b- 

B        (. 

ARGUMENT 

1.  Prolong  c  through  B  until 

prolongation  BD  =  a. 

2.  Draw  CD. 

3.  In   isosceles  A  BDC, 

/.D  =  Z/)C'#. 

4.  But  ZZ)CJ  >  ^DCB. 

5.  .-.  ZDCM  >  ZD. 

6.  /.  in  A  ADC,  AD  >  6. 


7.    .-.  a  -f  c  >  b. 


Q.E.D. 


REASONS 

1.  Str.  line  post.  II.    §  54, 16. 

2.  Str.  line  post.  I.     §  54,  15. 

3.  The  base  A  of  an  isosceles 

A  are  equal.     §  111. 
4.'  The   whole  >  any   of    its 
parts.     §  54,  12. 

5.  Substituting   /-D    for    its 

equal,  /.DCS. 

6.  If  two  A  of  a  A  are  unequal, 

the  side  opposite  the 
greater  Z  is  >  the  side 
opposite  the  less  angle. 
§164. 

7.  Substituting  a  -f-  c  for  AD. 


168.  Cor.  I.     Any  side  of  a  triangle  is  less  than  the  sum 
(did  greater  than  the  difference  of  the  other  two. 

169.  Cor.  II.     Any  straight  line  is  less  than  the  sum,  of 
the  parts  of  a  broken  line  having  the  same  extremities. 


60  PLANE   GEOMETRY 

170.  Note  to  Teacher.  Teachers  who  prefer  to  assume  that 
"  a  straight  line  is  the  shortest  line  between  two  points  "  may 
omit  Prop.  XVIII  entirely.  Then  Prop.  XVII  may  be  proved 
by  a  method  similar  to  that  used  in  Prop.  XV.  (See  Ex.  172.) 


Ex.  172.     Prove  Prop.  XVII  by  using  the  hint  contained  in  §  158. 

Ex.  173.  If  two  sides  of  a  triangle  are  14  and  9,  between  what  limit- 
ing values  must  the  third  side  be  ? 

Ex.  174.  If  the  opposite  ends  of  any  two  non-intersecting  line  seg- 
ments are  joined,  the  sum  of  the  joining  lines  is  greater  than  the  sum  of 
the  other  two  lines. 

Ex.  175.  Given  two  points,  P 
and  J?,  and  a  line  AB  not  passing 
through  either.  To  find  a  point  O, 
on  AB,  such  that  PO  +  OH  shall 
be  as  small  as  possible. 

This  exercise  illustrates  the  law 
by  which  light  is  reflected  from  a 
mirror.  The  light  from  the  object, 
P,  is  reflected  and  appears  to  come  L 

from  L,  as  far  behind  the  mirror  as  P  is  in  front  of  it. 

Ex.  176.  If  from  any  point  within  a  triangle  lines  are  drawn 
to  the  extremities  of  any  side  of  the  triangle,  the  sum  of  these  lines  is 
less  than  the  sum  of  the  other  two  sides  of  the  triangle. 

HINT.  Let  ABC  be  the  given  triangle,  D  the  point  within.  Prolong 
AD  until  it  intersects  BC  at  E.  Apply  Prop.  XVIII. 


171.  Note  to  Teacher.  Up  to  this  point  all  proofs  given  have 
been  complete,  including  argument  and  reasons.  In  written 
.work  it  is  frequently  convenient,  however,  to  have  students 
give  the  argument  only.  These  two  forms  will  be  distinguished 
by  calling  the  former  a  complete  demonstration  and  the  latter, 
which  is  illustrated  in  Prop.  XIX,  argument  only. 

It  is  often  a  sufficient  test  of  a  student's  understanding  of  a 
theorem  to  have  him  state  merely  the  main  points  involved  in 
a  proof.  This  may  be  given  in  enumerated  steps,  as  in  Prop. 
XXXV,  or  in  the  form  of  a  paragraph,  as  in  Prop.  XLIV.  This 
form  will  be  called  outline  of  proof. 


BOOK  I 


61 


PROPOSITION  XIX.     THEOREM 

172.  If  two  triangles  have  two  sides  of  one  equal  respec- 
tively to  two  sides  of  the  other,  but  the  included  angle  of 
the  first  greater  than  the  included  angle  of  the  second,  then 
the  third  side  of  the  first  is  greater  than  the  third  side  of 
the  second. 


F  /* 

Given    two  A  ABC  and  DEF  with  AB  =  DE,  BC  =  .RF,    but 


ARGUMENT  ONLY 


To  prove    AC  >  DF. 


1.  Place  A  D^F  on  A  ABC  so  that  DE  shall  fall  upon  its  equal 

AB,  D  upon  ^4,  E  upon  5. 

2.  J:F  will  then  fall  between  AB  and  BC.     Denote  A  DEF  in 

its  new  position  by  ABF. 

3.  Draw  BK  bisecting  /.  FBC  and  meeting  AC  at  JT. 

4.  Draw  .O*. 

5.  In  A  J^JT  and  -fiTtfC,  5(7  =  BF. 

6.  BK—BK. 

7.  Z.FBK—/.KBC. 

8.  .-.  AFBK=AKBC. 

9.  .'.KF=KC. 

10.  ^JT+  KF>AF. 

11.  .-.  ^^4-^(7>^^. 

12.  That  is,  ^C>^^. 

13.  .-.  AC>DF.  Q.E.D. 


62 


PLANE   GEOMETRY 


Ex.  177.  (a)  Draw  a  figure  and  discuss  the  case  for  Prop.  XIX 
when  F  falls  on  AC;  when  F  falls  within  triangle  ABC.  (6)  Discuss 
Prop.  XIX,  taking  AC  (the  base)  =  DF,  AB  =  DE,  and  angle  CAB 
greater  than  angle  D. 

Ex.  178.    Prove  Prop.  XIX  by  using  Fig.  1. 

HINT  .     In  A  A  CE,  Z  CEA  >  Z  BE  A. 

:.  Z  CEA  >  Z  EAB  >  Z  EA  C.  FIG.  1. 

Ex.  179.  Given  triangle  ABC  with  AB  greater  than  BC,  and  let 
point  P  be  taken  on  A  B  and  point  Q  on  <ZB,  so  that  AP  =  CQ.  Prove 
J.$  greater  than  CP. 

Ex.  180.  Would  the  conclusion  of  Ex. 
179  be  true  if  P  were  taken  on  AB  pro-  B 

longed  and  Q  on  CB  prolonged  ?     Prove. 

Ex.  181.  In  quadrilateral  ABCD  if 
AB  —  CD,  and  angle  CD  A  is  greater  than 
angle  DAB,  prove  AC  greater  than  BD. 


PROPOSITION  XX.     THEOREM 

(Converse  of  Prop.  XIX) 

173.  If  two  triangles  have  two  sides  of  one  equal  respec- 
tively to  two  sides  of  the  other,  but  the  third  side  of  the 
first  greater  than  the  third  side  of  the  second,  then  the 
angle  opposite  the  third  side  of  the  first  is  greater  than 
the  angle  opposite  the  third  side  of  tJie  second. 


B 


Given     A  ABC  and  DEF  with.  AB  =  DE,  BC  =  EF,  but  AC  >  DF. 
To  prove    /-B  >  ZlE. 


BOOK   I 


63 


ARGUMENT 

1.  /-B</.E,        Z.B  —  Z.E, 

or       /.B  >  Z#. 

2.  First  suppose  Z  B  <  Z  tf ; 

then  AC<DF. 


3.  This  is  impossible. 

4.  Next  suppose  ZJB  =  Z 

then  AABC  =  ADEF. 


5.  ..'.  AC—DF. 

6.  This  is  impossible. 

7.  .-.  Z.B>Z.E. 


Q.E.D. 


REASONS 

1.  In  this  case  only  three  sup- 

positions are  admissible. 

2.  If  two  A  have  two  sides  of 

one  equal  respectively  to 
two  sides  of  the  other, 
but  the  included  Z  of 
the  first  >  the  included 
Z  of  the  second,  then  the 
third  side  of  the  first  > 
the  third  side  of  the 
second.  §  172. 

3.  By  hyp.,  AC>DF. 

4.  Two  A  are  equal  if  twa 

sides  and  the  included  Z 
of  one  are  equal  respec- 
tively to  two  sides  and 
the  included  Z  of  the 
other.  §  107. 

5.  Homol.  parts  of  equal  fig- 

ures are  equal.     §  110. 

6.  Same  reason  as  3. 

7.  The  two  suppositions,  Z,B 

<  Z.E  and  Z£  =  ZJE:, 
have  been  proved  false. 


Ex.  182.  If  two  opposite  sides  of  a  quadrilateral  are  equal,  but  its 
diagonals  are  unequal,  then  one  angle  opposite  the  greater  diagonal  is 
greater  than  one  angle  opposite  the  less  diagonal. 

Ex.  183.  If  two  sides  of  a  triangle  are  unequal,  the  median  drawn  to 
the  third  side  makes  unequal  angles  with  the  third  side. 

Ex.  184.  If  from  the  vertex  8  of  an  isosceles  triangle  RST  a  line  is 
drawn  to  point  P  in  the  base  BT  so  that  HP  is  greater  than  P71,  then 
angle  BSP  is  greater  than  angle  PST. 

Ex.  185.  If  one  angle  of  a  triangle  is  equal  to  the  sum  of  the  other 
two,  the  triangle  can  be  divided  into  two  isosceles  triangles. 


64  PLANE   GEOMETRY 

SUMMARY    OF    THEOREMS    FOR   PROVING   ANGLES    UNEQUAL 

174.  (a)    When  the  angles  are  in  the  same  triangle  : 

If  two  sides  of  a  triangle  are  unequal,  the  angle  opposite  the 
greater  side  is  greater  than  the  angle  opposite  the  less  side. 

(&)    When  the  angles  are  in  different  triangles: 

If  two  triangles  have  two  sides  of  one  equal  respectively  to 
two  sides  of  the  other,  but  the  third  side  of  the  first  greater 
than  the  third  side  of  the  second,  then  the  angle  opposite  the 
third  side  of  the  first  is  greater  than  the  angle  opposite  the 
third  side  of  the  second. 

(c)  An  exterior  angle  of  a  triangle  is  greater  than  either  re- 
mote interior  angle. 

SUMMARY    OF    THEOREMS    FOR    PROVING   LINES    UNEQUAL 

175.  (a)    When  the  lines  are  in  the  same  triangle : 

The  sum  of  any  two  sides  of  a  triangle  is  greater  than  the 
third  side. 

Any  side  of  a  triangle  is  less  than  the  sum  and  greater  than 
the  difference  of  the  other  two. 

If  two  angles  of  a  triangle  are  unequal,  the  side  opposite  the 
greater  angle  is  greater  than  the  side  opposite  the  less  angle. 

(b)  When  the  lines  are  in  different  triangles  : 

If  two  triangles  have  two  sides  of  one  equal  respectively  to 
two  sides  of  the  other,  but  the  included  angle  of  the  first 
greater  than  the  included  angle  of  the  second,  then  the 
third  side  of  the  first  is  greater  than  the  third  side  of  the 
second. 

(c)  Every  point  not  in  the  perpendicular  bisector  of  a  line 
is  not  equidistant  from  the  ends  of  the  line. 

The  perpendicular  is  the  shortest  straight  line  from  a  point 
to  a  line.  • 

Any  straight  line  is  less  than  the  sum  of  the  parts  of  a 
broken  line  having  the  same  extremities. 

In  some  texts  Exs.  168  and  169  are  given  as  theorems. 


BOOK  I 


65 


176.  Note.  The  student  should  note  that  the  proof  of  Prop.  XIX 
depends  upon  the  device  of  substituting  a  broken  line  for  an  equal  straight 
line.  This  method  is  further  illustrated  by  the  following  exercises. 


B     E 


Ex.  186.  Prove  by  the  method  discussed 
above  that  lines  drawn  to  the  ends  of  a  line 
from  any  point  not  in  its  perpendicular 
bisector  are  unequal. 

Ex.  187.  Construct  an  isosceles  triangle, 
given  the  base  and  the  sum  of  the  altitude  and 
a  side. 

SOLUTION.  Let  b  be  the  required  base  and 
let  a  4-  s  be  the  sum  of  the  altitude  and  a  side. 
Imagine  the  problem  solved,  giving  A  ABC. 
By  marking  the  given  lines  and  studying  the 
figure,  the  following  procedure  will  be  found 
to  be  possible  :  On  an  unlimited  line  CE  lay 
off  CB  =  b  ;  next  draw  the  _L  bisector  of  CB 

and  upon  it  lay  off  FD  =  a  +  s.  It  is  now  necessary  to  break  off  a  part 
of  a  +  s  to  form  AB.  The  fact  that  AB  must  equal  AD  should  suggest  an 
isosceles  A  of  which  BD  will  be  the  base  and  of  which  it  is  required  to  find 
the  vertex.  But  the  _L  bisector  of  the  base  of  an  isosceles  A  passes 
through  the  vertex.  Therefore,  the  solution  is  completed  by  drawing  the  JL 
bisector  of  BD,  which  determines  A,  and  by  drawing  AB  and  AC. 

Ex.  188.     Construct  a  triangle,  given  the  base,  an  adjacent  angle,  and 
the  sum  of  the  other  two  sides. 

Ex.  189.     Construct  a  right  triangle,  given  one  arm  and  the  sum  of  the 
hypotenuse  and  the  other  arm. 


a  +  s 


PARALLEL   LINES 

177.  Def.    Two  lines  are  parallel  if  they  lie  in  the  same  plane 
and  do  not  meet  however  far  they  are  prolonged  either  way. 

178.  Assumption  19.     Parallel  line  postulate.     Two   intersect- 
ing straight  lines  cannot  both  be  parallel  to  the  same  straight  line. 

179.  The  following  form  of  this  postulate  is  sometimes  more 
convenient  to  quote  :   Tlirough  a  given  point  there  exists  only  one 
line  parallel  to  a  given  line. 


66 


PLANE   GEOMETRY 


PROPOSITION  XXI.     THEOREM 

180.   If  two  straight    lines    are    parallel    to  a    third 
straight  line,  they  are  parallel  to  each  other. 


Given   lines  a  and  b,  each  II  c. 
To  prove   a  II  b. 


ARGUMENT 

1.  a   and    b   are   either  II    or 

not  II. 

2.  Suppose  that  a  is  not  II  b ; 

then  they  will  meet  at 

some  point  as  D. 
3  This  is  impossible. 
4.  .-.  a  II  b. 

Q.E.D. 


REASONS 

1.  In  this  case  only  two  sup- 

positions are  admissible. 

2.  By  clef,  of  II  lines.     §  177. 


3.  Parallel  line  post.     §  178. 

4.  The  supposition  that  a  and 

b   are    not    II    has   been 
proved  false. 


181.  Def.     A  transversal  is  a  line  that   intersects   two  or 
more  other  lines. 

182.  Defs.     If  two  straight  lines  are  cut  by  a  transversal, 
of  the  eight  angles  formed, 

3,  4,  5,  6  are  interior  angles ; 

1,  2,  7,  8  are  exterior  angles ; 

4  and  5,  3  and  6,  are  alternate 
interior  angles ; 

1  and  8,  2  and  7,  are  alternate 
exterior  angles ; 

1  and  5,  3  and  7,  2  and  6,  4  and 
8,  are  corresponding  angles  (called  also  exterior  interior  angles). 


5\6 


•V 


BOOK  I 


67 


PROPOSITION  XXII.     THEOREM 

183.  If  two  straight  lines  are  cut  by  a  transversal 
making  a  pair  of  alternate  interior  angles  equal,  the 
lines  are  parallel. 


Given   two  str.  lines  MN  and  OQ  cut  by  the  transversal  AB  in 
points  C  and  D}  making  Z.  MCD  =  Z.QDC. 
To  prove    MN  II  OQ. 


ARGUMENT 


1. 


or 


MN  and  OQ  are  either 
not  II. 

2.  Suppose  that  MN  is  not  II  OQ ; 

then  they  will  meet  at 
some  point  as  P,  forming, 
with  line  DC,  A  PDC. 

3.  Then  /.MCD  >  /_  QDC. 


4.  This  is  impossible. 

5.  .-.  MN  II  OQ. 


Q.E.D. 


REASONS 

1.  In  this  case  only  two  sup- 

positions are  admissible. 

2.  By  def.  of  II  lines.     §  177. 


3.  If  one  side  of  a  A  is  pro- 

longed, the  ext.  /-  formed 
>  either  of  the  remote 
int.  A.  §  153. 

4.  Z.MCD  =  /.QDC,  by  hyp. 

5.  The  supposition  that  MN 

and  OQ  are  not  II  has  been 
proved  false. 


184.  Cor.  I.  If  two  straight  lines  are  cut  by  a  trans- 
versal making  a  pair  of  corresponding  angles  equal,  the 
lines  are  parallel. 

HINT.     Prove  a  pair  of  alt.  int.  A  equal,  and  apply  the  theorem. 


68 


PLANE   GEOMETRY 


185.  Cor.  n.     If  two  straight  lines  are  cut  by  a  trans- 
versal making  a  pair  of  alternate  exterior  angles  equal, 
the  lines  are  parallel.    (HINT.    Prove  a  pair  of  alt.  int.  A  equal.) 

186.  Cor.  HI.     If  two  straight  lines  are  cut  by  a  trans- 
versal making  the  suin  of  the  two  interior  angles  on  the 
same  side  of  the  transversal  equal  to  two  right  angles,  the 
lines  are  parallel. 


Given  lines  m  and  n  cut  by  the 
transversal  t  making 

Zl  +  Z2  =  2rt.Zs. 
To  prove    m  ||  n. 


m 


L 


I 


i 


2. 


3. 


4. 


5. 


ARGUMENT 
=  2rt,  A 
=  2rt.A 


Z2  =  Z3. 


m    n. 


Q.E.D. 


REASONS 

1.  By  hyp. 

2.  If  one  str.  line  meets  an- 

other str.  line,  the  sum 
of  the  two  adj.  A  is  2 
rt.  Zs.  §  65. 

3.  Things  equal  to  the  same 

thing  are  equal  to  each 
other.  §  54,  1. 

4.  If    equals    are    subtracted 

from  equals,  the  remain- 
ders are  equal.  §  54,  3. 

5.  If  two   str.  lines   are   cut 

by  a  transversal  making 
a  pair  of  alt.  int.  A  equal, 
the  lines  are  II.  §  183. 


187.   Cor.  IV.     If  two  straight  lines  are  perpendicular 
to  a  third  straight  line,  they  are  parallel  to  each  other. 


Ex.  190.  If  two  straight  lines  are  cut  by  a  transversal  making  the  sum 
of  the  two  exterior  angles  on  the  same  side  of  the  transversal  equal  to  two 
right  angles,  the  lines  are  parallel. 


BOOK  I  69 


Ex.  191.    In  the  annexed  diagram,  if   angle 


Prove. 


=  angle  CHF,  are  AS  and  CD  parallel?  V 

* 


Ex.  192.    In  the  same  diagram,  if  angle  HGB 


=  angle  GHC,  prove  that  the  bisectors  of  these  ^\r» 

angles  are  parallel. 

Ex.  193.   If  two  straight  lines  bisect  each  other,  the  lines  joining  their 
extremities  are  parallel  in  pairs. 

Ex.  194.    In  the  diagram  for  Ex.  191,  if  angle  BGE  and  angle  FHD 
are  supplementary,  prove  AB  parallel  to  CD. 

Ex.  195.    If  two  adjacent  angles  of  any  quadrilateral  are  supplemen- 
tary, two  sides  of  the  quadrilateral  will  be  parallel. 

PROPOSITION  XXIII.     PROBLEM 

188.   Through  a  given  point  to  construct  a  line  parallel 
to  a  given  line. 

8S 

, */vr 

.  »       ./  J 


/R         K 

Given   line  AB  and  point  P. 

To  construct,   through  point  P,  a  line  II  AB. 

I.    Construction 

1.  Draw  a  line  through  P  cutting  AB  at  some  point,  as  R. 

2.  With  P   as  vertex   and  PS  as    side,  construct  Z  YPS  = 
Z  BRP.     §  125. 

3.  XY  will  be  II  AB. 

II.     The  proof  and  discussion  are  left  as  an  exercise  for  the 
student. 

Ex.  196.  Through  a  given  point  construct  a  parallel  to  a  given  line  by 
using  :  (a)  §  183  ;  (b)  §  185  ;  (c)  §  187, 


70  PLANE    GEOMETRY 

PROPOSITION  XXIV.     THEOREM 
(Converse  of  Prop.  XXII) 

189.   If  two  parallel  lines  are  cut  ~by  a  transversal,  the 
alternate  interior  angles  are  equal. 

tE 


-Y 
-B 


X — 

c — 


Given    ||  lines  AB  and  CD  cut  by  the  transversal  EF  at  points 

and  H. 

To  prove    Z  A  GH  =  Z  DHG. 


ARGUMENT 

1.  Either   Z  A  Gil  =  Z  DHG, 

or  Z.AGH=f=  /.DHG. 

2.  Suppose  Z. 4  GH=^ZDHG, 

but  that  line  XF, 
through  G,  makes 
ZXGH  =  Z.DHG. 

3.  Then  .YF  II  C7Z>. 


4.  But  A  B  ||  CD. 

5.  It  is  impossible  that  AB 

and  ^r  both  are  ||  CD. 

6.  .'.  Z  AGH=  /.DHG. 

Q.K.D. 


REASONS 

1.  In  this  case  only  two  sup- 

positions are  admissible. 

2.  With  a  given  vertex  and 

a  given  side,  an  Z  may 
be  constructed  equal  to  a 
given  Z.  §  125. 

3.  If  two  str.  lines  are  cut  by 

a  transversal  making  a 
pair  of  alt.  int.  Z  equal, 
the  lines  are  || .  §183. 

4.  By  hyp. 

5.  Parallel  line  post.     §  178. 


6.  The  supposition  that 
Z  A  GH  =£  Z  DHG  has  been 
proved  false. 


BOOK  I  71 

190.  Cor.  I.    (Converse  of  Cor.  I  of  Prop.  XXII).     //  two 
parallel  lines  are  cut  by  a  transversal,  the  corresponding 
angles  are  equal. 

Given  two  ||  lines  XY  and  MN     % 
cut  by  the  transversal  AB,  form- 
ing corresponding  A  1  and  2.  ^  \a  N 
To  prove    Z  1  =  Z2.                                                            \ 
MINT.     /.  3  =  Z.  2  by  §  189.                                                           \fl 

191.  Cor.   II.    (Converse    of    Cor.    II     of    Prop.     XXII). 
If  two  parallel  lines  are  cut  by  a  transversal,  the  alter- 
nate exterior  angles  are  equal. 

192.  Cor.   m.    (Converse    of    Cor.    Ill    of    Prop.    XXII). 
If  two  parallel  lines  are  cut  by  a  transversal,  the  sum 
of  the  two  interior  angles  on  the  same  side  of  the  trans- 
versal is  two  right  angles.  * 

Given  two  II  lines  XY  and   MN    X 
cut  by  the  transversal  AB,  form- 
ing int.  A  1  and  2.  Af ^__ 

To  prove   Zl+Z2  =  2rt.  A.  / 

B/ 
ARGUMENT  ONLY 

Z3  =  Z2. 
:  1  -f  Z  2  =  2  rt.  A.  Q.E.D. 

193.  Cor.  IV.     A  straight  line  perpendicular  to  one  of 
two  parallels  is  perpendicular  to  the  other  also. 

194.  Cor.    V.      (Opposite    of    Cor.    Ill   of    Prop.    XXII); 
If  two  straight  lines  are  cut  by  a  transversal  making 
the  sum  of  the  two  interior  angles  on  the  same  side  of 
the  transversal  not  equal  to  two  right  angles,  the  lines 
are  not  parallel.     (HINT.    Apply  §  137,  or  use  the  indirect  method.) 


72  PLANE   GEOMETRY 

195.   Cor.  VI.     Two  lines  perpendicular  respectively  to 
two  intersecting  lines  also  intersect. 

Given  two  intersecting  lines  AB 
and  CD,  and  BE  J,  AB,  GF  _L  CD. 

To  prove  that  BE  and  CF  also 
intersect. 


ARGUMENT  ONLY 

1.  Draw  CB. 

2.  /.FCD  is  a  rt.  Z. 

3.  .-.  Zl  <  a  rt.  Z. 

4.  Likewise  Z2  <  a  rt.  Z. 

5.  .-.  Z1  +  Z2  <  2rt.  A. 

6.  .-.  BE  and  CF  also  intersect.  Q.E.D. 

196.   Def.    Two  or  more  lines  are  said  to  be  concurrent  if 

they  intersect  at  a  common  point. 


Ex.  197.  If  two  parallels  are  cut  by  a  transversal  so  that  one  of  the 
angles  formed  is  45°,  how  many  degrees  are  there  in  each  of  the  other 
seven  angles  ? 

Ex.  198.  .  If  a  quadrilateral  has  two  of  its  sides  parallel,  two  pairs  of 
its  angles  will  be  supplementary.  „  .-, 

Ex.  199.     In    the   annexed    diagram  AB   is  \         \ 

parallel    to    CD,    and  EF   is  parallel    to   GH.  A — V A^- B 

Prove  c        \2      \        ft 
(a)     angle  1  =  angle  2  ;  \ >^ 

(&)     angle  1  +  angle  3  =  2  right  angles.  F        H 

Ex.  200.  If  a  line  is  drawn  through  any  point  in  the  bisector  of  an 
angle,  parallel  to  one  of  the  sides  of  the  angle,  an  isosceles  triangle  will  be 
formed. 

Ex.  201.  Draw  a  line  parallel  to  the  base  of  a  triangle,  cutting  the 
sides  so  that  the  sum  of  the  segments  adjacent  to  the  base  shall  equal  the 
parallel  line. 

Ex.  202.  If  two  parallel  lines  are  cut  by  a  transversal,  the  bisectors 
of  a  pair  of  corresponding  angles  are  parallel. 


BOOK  I  73 

Ex.  203.     State  and  prove  the  converse  of  Ex.  190. 
Ex.  204.     State  and  prove  the  converse  of  Ex.  202. 

Ex.  205.  If  a  line  is  drawn  through  the  vertex  of  an  isosceles  triangle 
parallel  to  the  base,  this  Hue  bisects  the  exterior  angle  at  the  vertex. 

Ex.  206.     State  and  prove  the  converse  of  Ex.  205. 

Ex.  207.  If  a  line  joining  two  parallels  is  bisected,  any  other  line 
through  the  point  of  bisection  and  limited  by  the  parallels  is  bisected. 

Ex.  208.  If  through  the  vertex  of  one  of  the  acute  angles  of  a  right 
triangle  a  line  is  drawn  parallel  to  the  opposite  side,  the  line  forms 
with  the  hypotenuse  an  angle  equal  to  the  other  acute  angle  of  the 
triangle. 

Ex.  209.     The  acute  angles  of  a  right  triangle  are  complementary. 

Ex.  210.  If  two  sides  of  a  triangle  are  prolonged  their  own  lengths 
through  the  common  vertex,  the  line  joining  their  ends  is  parallel  to  the 
third  side  of  the  triangle. 

Ex.  211.  In  an  isosceles  triangle,  if  equal  segments  measured  from 
the  vertex  are  laid  off  on  the  arms,  the  line  joining  the  ends  of  the  seg- 
ments is  parallel  to  the  base  of  the  triangle.  (HINT.  Draw  the  bisector 
of  the  vertex  Z  of  the  A.) 

Ex.  212.  Extend  Ex.  211  to  the  case  where  the  segments  are  external 
to  the  triangle. 

SUMMARY    OF    THEOREMS    FOR    PROVING    LINES    PARALLEL 

alternate  interior  angles  equal, 


197.  (1)  If  two  straight 
lines  are  cut  by  a  trans- 
versal making  the 


alternate  exterior  angles  equal, 
corresponding  angles  equal, 
interior  angles  on  one  side  of  the 
transversal  supplementary, 

the  lines  are  parallel. 

(2)  Two  straight  lines  perpendicular  to  a  third  straight  line 
are  parallel  to  each  other. 

(3)  Two  straight  lines  parallel  to  a  third  straight  line  are 
parallel  to  each  other. 

(4)  After  parallelograms  have  been  studied,  the  fact  that 
the   opposite   sides  of   a  parallelogram  are  parallel   may   be 
used  (§  220). 


74 


PLANE   GEOMETRY 


PROPOSITION  XXV.     THEOREM 

198.    Two  angles  whose  sides  are  parallel,  each  to  each f 
are  either  equal  or  supplementary. 


Given    Zl  and  the  A  at  E,  with  AB  \\  EF  and  AC  ||  DE. 
To  prove   Z 1  ==  Z  2  =  Z3,  and   Z 1  +  Z  4  =  2  rt.  4  Zl 
5  =  2rt.Zs. 


..ARGUMENT 

1.  Prolong  £.4  and  DE  until 
they  intersect  at  some 
point  as  H. 


3.  Z2  =  Z6. 

4.  .-.  Z1  =  Z2. 


5.    Z3=Z2. 


6    .-.  Z1='Z3. 

7.   Z2      Z4  = 


8.    .-.  Z  1  +  Z  4  =  2  rt.  A 


REASONS 
1.    Str.linepost.il.    §54,16. 


2.  Corresponding  A  of  ||  lines 

are  equal.     §  190. 

3.  Same  reason  as  2. 

4.  Things  equal  to  the  same 

thing  are  equal  to  each 
other.  §  54,  1. 

5.  If  two  str.  lines  intersect, 

the  vertical  A  are  equal. 
§77. 

6.  Same  reason  as  4. 

7.  If  one  str.  line  meets  an- 

other str.  line,  the  sum 
of  the  two  adj.  A  is  2 
rt.  A.  §  65. 

8.  Substituting  Z 1   for   its 

equal  Z  2. 


BOOK  1 


75 


ARGUMENT 
9.   Z5=Z4. 
10.    .-.  Zl  +  Z5=  2rt.  A 

Q.E.D. 


REASONS 

9.    Same  reason  as  5. 
10.    Substituting   Z5   for 
equal  Z4. 


its 


199.  Note.  Every  angle  viewed  from  its  vertex  has  a  right  and  a  left 
side  ;  thus,  in  the  annexed  diagram,  the  right  side  of  Z.A  is  r  and  the  left 
side,  I ;  the  right  side  of  ZB  is  r  and  the  left,  I. 


FIG.  3. 


In  the  diagram  of  Prop.  XXV,  Zl  and  22,  whose  sides  are  ||  right  to 
right  (ABto  EF)  and  left  to  left  (AC  to  EK)  are  equal ;  while  2 1  and 
24,  whose  sides  are  ||  right  to  left  (AB  to  EF)  and  left  to  right  (AC  to 
ED)  are  supplementary.  Hence  : 

200.  (a)  If  two  angles  have  their  sides  parallel  right  to  right 
and  left  to  left,  they  are  equal. 

(b)  If  two  angles  have  their  sides  parallel  right  to  left  and  left 
to  right,  they  are  supplementary. 


Ex.  213.  In  Fig.  3,  above,  show  which  is  the  right  side  of  each  of  the 
four  angles  about  O. 

Ex.  214.  If  a  quadrilateral  has  its  opposite  sides  parallel,  its  opposite 
angles  are  equal. 

Ex.  215.  Given  two  equal  angles  having  a  side  of  one  parallel  to  a 
side  of  the  other,  are  the  other  sides  necessarily  parallel  ?  Prove. 

Ex.  216.  If  two  sides  of  a  triangle  are  parallel  respectively  to  two 
homologous  sides  of  an  equal  triangle,  the  third  side  of  the  first  is  parallel 
to  the  third  side  of  the  second. 

Ex.  217.  Construct  a  triangle,  given  an  angle  and  its  bisector  and 
the  altitude  drawn  from  the  vertex  of  the  given  angle. 


76 


PLANE   GEOMETRY 


PROPOSITION  XXVI.     THEOREM 

201.   Two  angles  whose  sides  are  perpendicular,  each  to 
each,  are  either  equal  or  supplementary. 

H         K 
E 


C         D 

Given   Zl  and  the  A  at  P,  with  OA±KC  and  OBj_HD. 

To  prove   Z1=Z2  =  Z3,  Zl+Z4=2rt.  Zs,Zl+Z5  =  2  rt.  A. 

HINT.     Draw  OE II  CK  and  OF  II  DH.     Prove  O#  J_  OA  and  OF  J_  OB. 
Prove  Z  7  =  Z  1,  and  prove  Z  7  =  Z  2. 

202.  Note.     It  will  be  seen  that  Zl  and  Z2,  whose  sides  are  _L  right 
to  right  (OA  to  PK}  and  left  to  left  (OS  to  P#),  are  equal ;  while  Zl 
and  Z4,  whose  sides  are  _L  right  to  left  (OA  to  PC)  and  left  to  right 
(OS  to  P£T),  are  supplementary.     Hence; 

203.  (a)  If  two  angles  have  their  sides  perpendicular  right  to 
right  and  left  to  left,  they  are  equal. 

(b)  If  two  angles  have  their  sides  perpendicular  right  to  left 
and  left  to  right,  they  are  supplementary. 


Ex.  218.  If  from  a  point  outside  of  an  angle  perpendiculars  are 
drawn  to  the  sides  of  the  angle,  an  angle  is  formed  which  is  equal  to  the 
given  angle. 

Ex.  219.  In  a  right  triangle  if  a  perpendicular  is  drawn  from  the 
vertex  of  the  right  angle  to  the  hypotenuse,  the  right  angle  is  divided 
into  two  angles  which  are  equal  respectively  to  the  acute  angles  of  the 
triangle. 

Ex.  220.  If  from  the  end  of  the  bisector  of  the  vertex  angle  of  an 
isosceles  triangle  a  perpendicular  is  dropped  upon  one  of  the  arms,  the 
perpendicular  forms,  witti  the  base  an  angle  equal  to  half  the  vertex  angle, 


BOOK   I 


77 


PROPOSITION  XXVII.     THEOREM 

204.    The  sum  of  the  angles  of  any  triangle  is  two  right 
angles. 


Given    A  ABC. 

To  prove    /.A  4-  Z  ABC  +  Z  C  =  2  rt.  A 

ARGUMENT  REASONS 

1.  Through  B  draw  D#  II  .4C.    .    1.   Parallel  line  post.     §  179. 

2.  Z  1  4-  Z  ABC  4-  Z  2  2.   The  sum  of  all  the  A  about 

=  2  rt.  Z.  a  point  on  one  side  of  a 

str.  line  passing  through 
that  point  =  2  rt.  A. 
§66. 

3.  Z1=Z^.  3.   Alt.  int.  A  of  II  lines  are 

equal.     §  189. 

4.  Z  2  =  Z  C.  4.    Same  reason  as  3. 

5.  .-.  /-A  4-  Z^.BC'  -f  ZC7  5.    Substituting  for  A  1  and  2 

their  equals,  Z  A  and  (7, 
respectively. 

205.  Cor.  I.     In  a  right  triangle  the  two  acute  angles 
are  complementary. 

206.  Cor.  II.     In  a  triangle  there  can  be  but  one  right 
angle  or  one  obtuse  angle. 

207.  Cor.  III.     If  two  angles  of  one  triangle  are  equal 
respectively  to  two  angles  of  another,  tlien  the  third  angle 
of  the  first  is  equal  to  the  third  angle  of  the  second. 

208.  Cor.  IV.     //  two   right   triangles   have   an  acute 
angle  of  one  equal  to  an  acute  angle  of  the  other,  the 
other  acute  angles  arc  equal. 


Z^  -f  /-ABC  4-  ZC7 

=  2  rt.  A. 

Q.E.D. 


78  PLANE   GEOMETRY 

209.  Cor.  V.   Two  right  triangles  are  equal  if  the  hy- 
potenuse and  an  acute  angle  of  one  are  equal  respectively 
to  the  hypotenuse  and  an  acute  angle  of  the  other. 

210.  Cor.  VI.    Two  right  triangles  are  equal  if  a  side 
and  an  acute  angle  of  one  are  equal  respectively  to  a 
side  and  the  homologous  acute  angle  of  the  other. 

211.  Cor.  VII.    Two  right  triangles   are  equal  if  the 
hypotenuse  and  a  side  of  one  are  equal  respectively  to  tlie 
hypotenuse  and  a  side  of  the  other. 

B 


K  A 


Given   rt.  A  ABC  and  DEF,  with  AB  =  DE  and  BC=EF. 

To  prove    A  ABC  =  A  DEF. 

HINT.     Prove  DFKa,  str.  line  ;  then  A  DEK  is  isosceles. 

212.  Cor.  VIII.    The  altitude  upon  the  base  of  an  isosceles 
triangle  bisects  the  base  and  also  the  vertex  angle. 

213.  Cor.  IX.    Each   angle   of  an  equilateral  triangle 
is  one  third  of  two  right  angles,  or  60°. 

214.  Question.     Why  is  the  word  homologous  used  in  Cor.  VI  but 
not  in  Cor.  V  ? 


Ex.  221.     If  any  angle  of  an  isosceles  triangle  is  60°,  what  is  the  value 
of  each  of  the  two  remaining  angles  ? 

Ex.  222.     If  the  vertex  angle  of  an  isosceles  triangle  is  20°,  find  the 
angle  included  by  the  bisectors  of  the  .base  angles. 

Ex.  223.     Find  each  angle  of  a  triangle  if  the  second  angle  equals 
twice  the  first  and  the  third  equals  three  times  the  second. 

Ex.  224.     If  one  angle  of  a  triangle  is  w°  and  another  angle  Z°,  write 
an  expression  for  the  third  angle. 

Ex.  225.     If  the  vertex  angle  of  an  isosceles  triangle  is  «°,  write  an 
expression  for  each  base  angle. 


BOOK  I 


79 


Ex.  226.     Construct  an  angle  of  60° ;  120°  ;  30°  ;  15°. 
Ex.  227.     Construct  a  right  triangle  having  one  of  its  acute  angles  60°. 
How  large  is  the  other  acute  angle  ? 
Ex.  228.     Construct  an  angle  of  150°. 


Ex.  229. 

above. 
Ex.  230. 
Ex.  231. 
Ex.  232. 


Trove  Prop.  XXVII  by  using  each  of  the  diagrams  given 


Given  two  angles  of  a  triangle,  construct  the  third  angle. 
Find  the  sum  of  the  angles  of  a  quadrilateral. 
The  angle  between  the  bisectors  of  two  adjacent  angles  of 
a  quadrilateral  is  equal  to  half  the  sum  of  the  two  remaining  angles. 

Ex.  233.  The  angle  between  the  bisectors  of  the  base  angles  of  an  isos- 
celes triangle  is  equal  to  the  exterior  angle  formed  by  prolonging  the  base. 

Ex.  234.  If  two  straight  lines  are  cut  by  a  transversal  and  the  bisectors 
of  two  interior  angles  on  the  same  side  of  the  transversal  are  perpendicular 
to  each  other,  the  lines  are  parallel. 

Ex.  235.     If  in  an  isosceles  triangle  each  of  the  base  angles  is  one 
fourth  the  angle  at  the  vertex,  a  line  drawn  perpendicular  to  the  base  at 
one  of  its  ends  and  meeting 
the  opposite  side  prolonged 
will  form  with  the  adjacent 
side  and  the  exterior  portion 
of    the     opposite     side    an 
equilateral  triangle.  A^ 

Ex.  236.      Two     angles 

whose  sides  are  perpendicular  each  to  each  are  either  equal  or  supple- 
mentary.    (Prove  by  using  the  annexed  diagram.) 

Ex.  237.  If  at  the  ends  of  the  hypotenuse  of  a  right  triangle  per- 
pendiculars to  the  hypotenuse  are  drawn  meeting  the  other  two  sides  of 
the  triangle  prolonged,  then  the  figure  contains  five  triangles  which  are 
mutually  equiangular. 

Ex.  238.  If  one  angle  of  a  triangle  is  50°  and  another  angle  is  70°, 
find  the  other  interior  angle  of  the  triangle ;  also  the  exterior  angles  of 
the  triangle.  What  relation  is  there  between  an  exterior  angle  and  the 
two  remote  interior  angles  of  the  triangle  ? 


B' 


B    B'  B 


80  PLANE    GEOMETRY 


PROPOSITION   XXVIII.     THEOREM 

215.   *dn  exterior  angle  of  a  triangle  is  equal  to  the  sum 
of  the  two  remote  interior  angles. 


C 

Given   AAEC  with  Zz><7£  an  exterior  Z. 

To  prove    ZDCB=Z.A  +  ^B. 

The  proof  is  left  as  an  exercise  for  the  student. 

HINT.     Draw  CE  II  AB. 


Ex.  239.  The  bisector  of  an  exterior  angle  at  the  vertex  of  an 
isosceles  triangle  is  parallel  to  the  base. 

Ex.  240.  If  the  sum  of  two  exterior  angles  of  a  triangle  is  equal  to 
three  right  angles,  the  triangle  is  a  right  triangle. 

Ex.  241.  The  sum  of  the  three  exterior  angles  of  a  triangle  is  four 
right  angles. 

Ex.  242.     What  is  the  sum  of  the  exterior  angles  of  a  quadrilateral  ? 

Ex.  243.  If  the  two  exterior  angles  at  the  base  of  any  triangle  are 
bisected,  the  angle  between  these  bisectors  is  equal  to  half  the  sum  of  the 
interior  base  angles  of  the  triangle. 

Ex.  244.  If  BE  bisects  angle  B  of 
triangle  ABC,  and  AE  bisects  the  exterior 
angle  DA  (7,  angle  E  is  equal  to  one  half 
angle  C. 

Ex.  245.  D  is  any  point  in  the  base 
BC  of  isosceles  triangle  ABC.  The  side  B  ~A  D 

AC  is  prolonged  through  C  to  E  so  that 

CE  =  CD,   and  DE  is  drawn  meeting  AB  at  F.    Prove   angle  EFA 
equal  to  three  times  angle  AEF. 

Ex.  246.  The  mid-point  of  the  hypotenuse  of  a  right  triangle  is 
equidistant  from  the  three  vertices.  Prove  by  laying  off  on  the  right 
angle  either  acute  angle. 


BOOK  I 


81 


PROPOSITION   XXIX.     THEOREM 

216.  The  sum  of  all  the  angles  of  any  polygon  is  twice 
as  many  right  angles  as  the  polygon  has  sides,  less  four 
right  angles. 


Given   polygon  ABODE  •  -  •,  any  polygon  having  n  sides. 
To  prove   the  sum  of  its  A  =  2  n  rt.  A  —  4  rt.  A. 


ARGUMENT 

1.  From  any  point  within  the 

polygon  such  as  0,  draw 
lines  to  the  vertices. 

2.  There  will  be  formed  n  A. 


3.  The  sum  of  the  A  of  each 

A  thus  formed  =  2  rt.  A. 

4.  .%  the  sum  of  the  A  of  the 

n  A  thus  formed  =  2  n 
rt.  A. 

5.  The    sum    of    all    the   A 

about  0  =  4  rt.  A. 

6.  .-.  the  sum  of  all  the  A  of 

the  polygon  =  2  n  rt.  A 
-  4  rt.  Zs. 

Q.E.D. 


REASONS 

1.    Straight  line  post  I. 
15. 


§54, 


2.  Each  side  of  the  polygon 

will  become  the  base  of 
a  A. 

3.  The  sum  of  the  A  of  any 

A  is  2  rt,  Zs.     §  204. 

4.  If    equals    are   multiplied 

by  equals,  the  products 
are  equal.  §  54,  7  a. 

5.  The  sum  of  all  the  A  about 

a  point  =  4  rt.  A.     §  67. 

6.  The  sum  of   all  the  A  of 

the  polygon  =  the  sum 
of  the  A  of  the  n  A  - 
the  sum  of  all  the  A 
about  O. 


217.   Cor.     Each  angle  of  an  equiangular  polygon  of 


n  sides  is  equal  to 


2(n  -  2) 


right  angles. 


82 


PLANE   GEOMETRY 


PROPOSITION  XXX.     THEOREM 

218.  //  the  sides  of  any  polygon  are  prolonged  in  succes- 
sion one  way,  no  two  adjacent  sides  being  prolonged 
through  the  same  vertex,  the  sum  of  the  exterior  angles 
thus  formed  is  four  right  angles. 


4/7) 


Given  polygon  P  with  Z 1,  Z  2,  Z  3,  Z4,  •  •  •  its  successive  ex- 
terior angles. 

To  prove   Z 1  +  Z  2  -f  Z3  +  ^4  H =  4  rt.  A. 


ARGUMENT 

1.  Z1  +  Z6  =  2  rt.   Zs,  Z2  + 

Z7  =  2  rt.  A,  and  so  on  ; 
i.e.  the  sum  of  the  int. 
Z  and  th'e  ext.  Z  at  one 
vertex  =  2  rt.  A. 

2.  .*.  the  sum  of  the  int.  and 

ext.  zS  at  the  n  vertices 
=  2  n  rt.  Z. 

3.  Denote  the  sum  of  all  the 

interior  A  by  /  and  the 
sum  of  all  the  ext.  A  by 
E-,  then  #-)-/=  2  nrt.  A. 

4.  But  l=2wrt.  Z-4rt.  A 


5.         .-.  J£  =  4  rt.  Z;    i.e.    Zl 

+  Z2  +  Z3+Z4+-  ... 
=  4  rt.  A  Q.E.D. 


REASONS 

1.  If  one  str.  line  meets  an- 

other str.  line,  the  sum  of 
the  two  adj.  A  is  2  rt.  A. 
§  65. 

2.  If  equals  are  multiplied  by 

equals,  the  products  are 
equal.  §  54,  7  a. 

3.  Arg.  2. 


4.  The  sum  of  all  the  A  of 

any  polygon  =  2  n  rt.  A 
-4rt.  Zs.     §216. 

5.  If    equals    are   subtracted 

from  equals,  the  remain- 
ders are  equal.     §  54,  3. 


BOOK  I  88 

219.    Note.       The  formula  2  n  rt.  A  -  4  rt.  A  (§  216)  is  sometimes 
more  useful  in  the  form  (;i  —  2)  2  rt.  A. 


Ex.  247.  Find  the  sum  of  the  angles  of  a  polygon  of  7  sides  ;  of  8 
sides ;  of  10  sides. 

Ex.  248.  Prove  Prop.  XXIX  by  drawing  as  many  diagonals  as 
possible  from  one  vertex. 

Ex.  249.     How  many  diagonals  can  be  drawn  from  one  vertex  in  a 
polygon  of  8  sides  ?  of  50  sides  ?  of  n  sides  ?     Show  that  the  greatest  num- 
ber of  diagonals  possible  in  a  polygon  of  n  sides  (using  all  vertices)  is 
n(n  -  3) 
2 

Ex.  250.  How  many  degrees  are  there  in  each  angle  of  an  equi- 
angular quadrilateral  ?  in  each  angle  of  an  equiangular  pentagon  ? 

Ex.  251.  How  many  sides  has  a  polygon  the  sum  of  whose  angles  is 
11  right  angles  ?  20  right  angles  ?  540°  ? 

Ex.  252.  How  many  sides  has  a  polygon  the  sum  of  whose  interior 
angles  is  double  the  sum  of  its  exterior  angles  ? 

Ex.  253.  Is  it  possible  for  an  exterior  angle  of  an  equiangular'poly- 
gon  to  be  70°  ?  72°  ?  140°  ?  144°  ? 

Ex.  254.  How  many  sides  has  a  polygon  each  of  whose  exterior 
angles  equals  12°  ? 

Ex.  255.  How  many  sides  has  a  polygon  each  of  whose  exterior 
angles  is  one  eleventh  of  its  adjacent  interior  angle  ? 

Ex.  256.  How  many  sides  has  a  polygon  the  sum  of  whose  interior 
angles  is  six  times  the  sum  of  its  exterior  angles  ? 

Ex.  257.  How  many  sides  has  an  equiangular  polygon  if  the  sum  of 
three  of  its  exterior  angles  is  180°  ? 

Ex.  258.  Tell  what  equiangular  polygons  can  be  put  together  to 
make  a  pavement.  How  many  equiangular  triangles  must  be  placed 
with  a  common  vertex  to  fill  the  angular  magnitude  around  a  point  ? 


TJL7 


84  PLANE   GEOMETRY 

QUADRILATERALS. '  PARALLELOGRAMS 

QUADRILATERALS    CLASSIFIED    WITH    RESPECT    TO    PARALLELISM 

220.  Def.     A  parallelogram  is  a  quadrilateral  whose  oppo- 
site sides  are  parallel. 

221.  Def.     A  trapezoid  is  a  quadrilateral  having  two  of  its 
opposite  sides  parallel  and  the  other  two  not  parallel. 

222.  Def.     A  trapezium  is  a  quadrilateral  having  no  two  of 
its  sides  parallel. 

PARALLELOGRAMS    CLASSIFIED    WITH    RESPECT   TO    ANGLES 

223.  Def.     A  rectangle  is  a  parallelogram  having  one  right 
angle. 

It  is  shown  later  that  all  the  angles  of  a  rectangle  are  right 
angles. 

224.  Def.     A  rhomboid  is  a  parallelogram  having  an  oblique 
angte. 

It  is  shown  later  that  all  the  angles  of  a  rhomboid  are 
oblique. 

225.  Def.     A  rectangle  having  two  adjacent  sides  equal  is 
a  square. 

It  is  shown  later  that  all  the  sides  of  a  square  are  equal. 

226.  Def.     A  rhomboid  having  two  adjacent  sides  equal  is 
a  rhombus. 

It  is  shown  later  that  all  the  sides  of  a  rhombus  are  equal. 

227.  Def.     A  trapezoid  having   its   two  non-parallel    sides 
equal  is  an  isosceles  trapezoid. 

228.  Def.     Any  side  of  a  parallelogram  may  be  regarded  as 
its  base,  and  the  line  drawn  perpendicular  to  the  base  from 
any  point  in  the  opposite  side  is  then  the  altitude. 

229.  Def.     The  bases  of  a  trapezoid  are  its  parallel  sides, 
and  its  altitude  is  a  line  drawn  from  any  point  in  one  base  per- 
pendicular to  the  other. 


BOOK   I 
PROPOSITION  XXXI.     THEOREM 


85 


230.   Any  two  opposite  angles  of  a  parallelogram,  are 
equal,  and  any  two  consecutive  angles  are  supplementary. 


Given    O  ABCD. 

To  prove :    (a)  /.A  =  Z (7,  and  /.B  =  /.D  ; 

(b)  any  two  consecutive  A,  as  A  and  B,  sup. 


ARGUMENT 
=  /.  Cand  Z.B=Z,D. 


A  A  and  B  are  sup. 


Q.E.D. 


REASONS 

1.  If  two  A  have  their  sides 

II  right  to  right  and  left 
to  left,  they  are  equal. 
§  200,  a. 

2.  If  two  II  lines-  are  cut  by  a 

transversal,  the  sum  of 
the  two  int.  A  on  the  same 
side  of  the  transversal  is 
two  rt.  A.  §  192. 


231.  Cor.    All  the  angles  of  a  rectangle  are  right  angles, 
and  all  the  angles  of  a  rhomboid  are  oblique  angles. 


Ex.  259. .  If  the  opposite  angles  of  a  quadrilateral  are  equal,  the  figure 
is  a  parallelogram. 

Ex.  260.     If  an  angle  of  one  parallelogram  is  equal  to  an  angle  of 
another,  the  remaining  angles  are  equal  each  to  each. 

Ex.  261.    The  bisectors  of  the  angles  of  a  parallelogram  (not  a  rhom- 
bus or  a  square)  inclose  a  rectangle, 


86  PLANE   GEOMETRY 


PROPOSITION  XXXII.     THEOREM 
232.     The  opposite  sides  of  a  parallelogram  are  equal. 


Given    O  ABCD. 

To  prove    AB  =  CD  and  BC  =  AD. 

The  proof  is  left  as  an  exercise  for  the  student. 

233.  Cor.  I.    All  the  sides  of  a  square  are  equal,  and 
all  the  sides  of  a  rhombus  are  equal. 

234.  Cor.  II.    Parallel   lines   intercepted   between  the 
same  parallel  lines  are  equal. 

235.  Cor.  III.    The  perpendiculars  drawn  to  one  of  two 
parallel  lines  from  any  two  points  in  the  other  are  equal. 

236.  Cor.  IV.    A  diagonal  of  a  parallelogram  divides 
it  into  two  equal  triangles. 


Ex.  262.  The  perpendiculars  drawn  to  a  diagonal  of  a  parallelogram 
from  the  opposite  vertices  are  equal. 

Ex.  263.  The  diagonals  of  a  rhombus  are  perpendicular  to  each  other 
and  so  are  the  diagonals  of  a  square. 

Ex.  264.     The  diagonals  of  a  rectangle  are  equal. 

Ex.  265.     The  diagonals  of  a  rhomboid  are  unequal. 

Ex.  266.  If  the  diagonals  of  a  parallelogram  are  equal,  the  figure  is  a 
rectangle. 

Ex.  267.  If  the  diagonals  of  a  parallelogram  are  not  equal,  the  figure 
is  a  rhomboid. 

Ex.  268.  Draw  a  line  parallel  to  the  base  of  a  triangle  so  that  the 
portion  intercepted  between  the  sides  may  be  equal  to  a  given  line. 

Ex.  269.  Explain  the  statement :  Parallel  lines  are  everywhere  equi- 
distant. Has  this  been  proved  ? 


BOOK  I 


87 


Ex.  270.  Find  the  locus  of  a  point  that  is  equidistant  from  two  given 
parallel  lines. 

Ex.  271.  Find  the  locus  of  a  point :  (a)  one  inch  above  a  given 
horizontal  line  ;  (&)  two  inches  below  the  given  line. 

Ex.  272.  Find  the  locus  of  a  point:  (a)  one  inch  to  the  right  of  a 
given  vertical  line  ;  (6)  one  inch  to  the  left  of  the  given  line. 

Ex.  273.  Given  a  horizontal  line  OX  and  a  line  O  Y  perpendicular 
to  OX.  Find  the  locus  of  a  point  three  inches  above  OJTand  two  inches 
to  the  right  of  O  Y. 


237.  Historical  Note.  Rend  Descartes  (1596-1650)  was  the  first  to 
observe  the  importance  of  the  fact  that  the  position  of  a  point  in  a  plane 
is  determined  if  its  distances, 
say  x  and  y,  from  two  fixed 
lines  in  the  plane,  perpendic- 
ular to  each  other,  are  known, 
lie  showed  that  geometric  fig- 
ures can  be  represented  by 
algebraic  equations,  and  de- 
veloped the  subject  of  analytic 
geometry,  which  is  known  by 
his  name  as  Cartesian  geome- 
try. 

Descartes  was  born  near 
Tours  in  France,  and  was  sent 
at  eight  years  of  age  to  the 
famous  Jesuit  school  at  La 
Fleche.  He  was  of  good  fam- 
ily, and  since,  at  that  time, 
most  men  of  position  entered  either  the  church  or  the  army,  he  chose  the 
latter,  and  joined  the  army  of  the  Prince  of  Orange. 

One  day,  while  walking  in  a  street  in  a  Holland  town,  he  saw  a  placard 
which  challenged  every  one  who  read  it  to  solve  a  certain  geometric 
problem.  Descartes  solved  it  with  little  difficulty  and  is  said  to  have 
realized  then  that  he  had  no  taste  for  military  life.  He  soon  resigned  his 
commission  and  spent  five  years  in  travel  and  study.  After  this  he  lived 
a  short  time  in  Paris,  but  soon  retired  to  Holland,  where  he  lived  for 
twenty  years,  devoting  his  time  to  mathematics,  philosophy,  astronomy, 
and  physics.  His  work  in  philosophy  was  of  such  importance  as  to  give 
him  the  name  of  the  Father  of  Modern  Philosophy. 


DESCARTES 


88  PLANE   GEOMETRY 


PROPOSITION  XXXIII.     THEOREM 

238.     The  diagonals   of  a  parallelogram  bisect  each 
other. 


A  D 

Given   O  ABCD  with  its  diagonals  AC  and  BD  intersecting  at  0. 
To  prove    AO  =  OC  and  BO  =  OD. 

HINT.     Prove  A  OBC  =  A  ODA.     .-.  AO  =  OC  and  BO  =  OD. 


Ex.  274.  If  through  the  vertices  of  a  triangle  lines  are  drawn  parallel 
to  the  opposite  sides  of  the  triangle,  the  lines  which  join  the  vertices  of 
the  triangle  thus  formed  to  the  opposite  vertices  of  the  given  triangle  are 
bisected  by  the  sides  of  the  given  triangle. 

Ex.  275.  A  line  terminated  by  the  sides  of  a  parallelogram  and  passing 
through  the  point  of  intersection  of  its  diagonals  is  bisected  at  that  point. 

Ex.  276.  How  many  parallelograms  can  be  constructed  having  a  given 
base  and  altitude  ?  What  is  the  locus  of  the  point  of  intersection  of  the 
diagonals  of  all  these  parallelograms  ? 

Ex.  277.  If  the  diagonals  of  a  parallelogram  are  perpendicular  to 
each  other,  the  figure  is  a  rhombus  or  a  square. 

Ex.  278.  If  the  diagonals  of  a  parallelogram  bisect  the  angles  of  the 
parallelogram,  the  figure  is  a  rhombus  or  a  square. 

Ex.  279.  Find  on  one  side  of  a  triangle  the  point  from  which  straight 
lines  drawn  parallel  to  the  other  two  sides,  and  terminated  by  those  sides, 
are  equal.  (See  §232.) 

Ex.  280.  Find  the  locus  of  a  point  at  a  given  distance  from  a  given 
finite  line  AB. 

Ex.  281.  Find  the  locus' of  a  point  at  a  given  distance  from  a  given 
line  and  also  equidistant  from  the  ends  of  another  given  line. 

Ex.  282.  Construct  a  parallelogram,  given  a  side,  a  diagonal,  and 
the  altitude  upon  the  pven  side. 


BOOK   I 


89 


PROPOSITION  XXXI V.     THEOREM 

(Converse  of  Prop.  XXXII) 

239.     If  the  opposite  sides  of  a  quadrilateral  are  equal, 
the  figure  is  a  parallelogram. 


D 


6. 

7. 


.'.  AB  II  CD. 


8.  Likewise  /.EDA  — 

9.  /.  BC  II  AD. 
10.  .-.  ABCD  is  a  O. 


BD. 

1.    S 

2.   B 

3.   B 

4.   B 

X 

5.    T 

6.   H 

7.   IJ 

8.    S 

9.    S 

Q.E.D. 

10.    B 

REASONS 

1.    Str.  line  post.  I.     §  54, 15. 
By  hyp. 


Given   quadrilateral  ABCD,  with  AB  =  CD,  and  BC=AD 
To  prove    ^LB  CD  a  O. 
ARGUMENT 

1.  Draw  the  diagonal  BD. 

2.  In  A  ABD  and 

AB  =  CD. 

3.  BC  =  ^D. 

4.  £D  =  BD. 

.-.  A  ABD  =  A  BCD.  5.  Two  A  are  equal  if  the 

three  sides  of  one  are 
equal  respectively  to 
the  three  sides  of  the 
other.  §  116. 

Homol.  parts  of  equal  fig- 
ures are  equal.  §  110. 

If  two  str.  lines  are  cut  by 
a  transversal  making  a 
pair  of  alt.  int.  A  equal, 
the  lines  are  II.  §  183. 

Same  reason  as  6. 

Same  reason  as  7. 


By  def.  of  a  O.    §  220. 


Ex.  283.     Construct  a  parallelogram,  given  two  adjacent  sides  and 
the  included  anirle. 


90  PLANE   GEOMETRY 

Ex.  284.     Construct  a  rectangle,  given  the  base  and  the  altitude. 
Ex.  285.     Construct  a  square,  given  a  side. 

Ex.  286.  Through  a  given  point  construct  a  parallel  to  a  given  line 
by  means  of  Prop.  XXXIV. 

Ex.  287.  Construct  a  median  of  a  triangle  by  means  of  a  parallelo- 
gram, (1)  using  §§  239  and  238  ;  (2)  using  §§  220  and  238. 

Ex.  288.  An  angle  of  a  triangle  is  right,  acute,  or  obtuse  according 
as  the  median  drawn  from  i.ts  vertex  is  equal  to,  greater  than,  or  less  than 
half  the  side  it  bisects. 

PROPOSITION  XXXY.    THEOREM 

240.  If  two  opposite  sides  of  a  quadrilateral  are  equal 
and  parallel,  the  figure  is  a  parallelogram. 


A  D 

Given    quadrilateral  ABCD,  with  BC  both  equal  and  II  to  AD. 
To  prove    ABCD  a  O. 

OUTLINE  OP  PROOF 

1.  Draw  diagonal  BD. 

2.  Prove  A  BCD  =  A  ABD. 

3.  Then  ZCDB  =  /.ABD  and  AB  II  CD. 

4.  .'.  ABCD  is  a  O. 


Ex.  289.     If  the  mid-points  of  two  opposite  sides  of  a  parallelogram 
are  joined  to  a  pair  of  opposite  vertices,  a  parallelogram  will  be  formed. 

Ex.  290.     Construct  a  parallelogram,  having  given  a  base,  an  adjacent 
angle,  and  the  altitude,  making  your  construction  depend  upon  §  240. 

Ex.  291.     If  the  perpendiculars  to  a  line  from  any  two  points  in  an- 
other line  are  equal,  then  the  lines  are  parallel. 

Ex.  292.     If  two  parallelograms  have  two  vertices  and  a  diagonal  in 
common,  the  lines  joining  the  other  four  vertices  form  a  parallelogram. 


BOOK  I  91 

PROPOSITION  XXXVI.     THEOREM 
(Converse  of  Prop.  XXXIII) 

241.   If  the  diagonals  of  a  quadrilateral  bisect  earh 
other,  the  figure  is  a  parallelogram. 

B  C 


A  D 

Given   quadrilateral  ABCD  with  its  diagonals  AC  and  BD  in- 
tersecting at  0  so  that  A0=  CO  and  BO  =  DO. 
To  prove    ABCD  a  O. 

ARGUMENT  ONLY 

1.  In  A  OBC  and  ODA, 

BO  =  DO. 

2.  GO  —  AO. 

3.  Z3  =  Z4. 

4.  .-.A  OBC=  A  OD^. 

5.  .'.  BC  =  AD. 

6.  Also  Zl  =  Z2. 

7.  .'.  5(71!  AD. 

8.  .-.  ABOD  is  a  O.  Q.E.D. 


Ex.  293.  In  parallelogram  ABCD,  let  diagonal  AC  be  prolonged 
through  ^4  and  C  to  X  and  T,  respectively,  making  ^1X=  OF.  Prove 
XB  YD  a  parallelogram. 

Ex.  294.  If  each  half  of  each  diagonal  of  a  parallelogram  is  bisected, 
the  lines  joining  the  points  of  bisection  form  a  parallelogram. 

Ex.  295.  Lines  drawn  from  the  vertices  of  two  angles  of  a  triangle 
and  terminating  in  the  opposite  sides  cannot  bisect  each  other. 

Ex.  296.  State  four  independent  hypotheses  which  would  lead  to  the 
conclusion,  "the  quadrilateral  is  a  parallelogram." 

Ex.  297.  Construct  a  parallelogram,  given  its  diagonals  and  an  angle 
between  them. 


92 


PLANE   GEOMETRY 


PROPOSITIOX  XXXVII.     THEOREM 

242.  Two  parallelograms  are  equal  if  two  sides  and  the 
included  angle  of  one  are  equal  respectively  to  two  sides 
and  the  included  angle  of  the  other. 


II 


Given   UJ  I  and  II  with  AB  —  EF,  AD  =  EH,  and  /.A=  /.E. 
To  prove    O.I  =  OIL 


ARGUMENT 

1.  Place  O I  upon  Oil  so  that 

AB  shall  fall  upon  its 
equal  EF,  A  upon  E,  B 
upon  F. 

2.  Then  AD  will  become  col- 

linear with  EH. 

3.  Point  D  will  fall  on  H. 

4.  Now  DC  II  AB,  and  #G  II  EF. 

5.  .-.  DC7  and  #£  are  both  II  AB. 

6.  .*.  DC  will  become  collinear 

with  #(?,  and  (7  will  fall 
somewhere  on  HG  or  its 
prolongation. 

7.  Likewise  ^(7  will   become 

collinear  with  FG,  and  C 
will  fall  somewhere  on 
FG  or  its  prolongation. 

8.  .-.  point    C   must    fall    on 

point  G. 

9.  ,-,  £71  =.OII,  Q.E.J>. 


REASONS 
1.   Transference  post.  §54,14. 


2.  Z^l  =  Z.E,  by  hyp. 

3.  ^4Z>  =  EH,  by  hyp. 

4.  By  def .  of  a  O.     §  220. 

5.  AB  and  EF  coincide,  Arg.  1. 

6.  Parallel  line  post.     §  179. 


7.   By  steps   similar   to  4,  5, 
and  6. 


8.  Two  intersecting  str.  lines 

can  have  only  one  point 
in  common.     §  26. 

9.  By  def.  of  equal  figures,  §  18. 


BOOK  I 


93 


243.    Quadrilaterals 


QUADRILATERALS    CLASSIFIED 

1.  Opposite  sides  II :  Parallelogram. - 

(a)    Right-angled:  Rectangle. 
Two  adj.  sides  equal: 

Square. 

(6)    Oblique-angled:  Rhomboid. 
Two  adj.  sides  equal : 
Rhombus. 

2.  Two  sides  II,  other  two  non-11 :   Trap- 

ezoid. 

(a)    Two  non-11  sides  equal : 
Isosceles  trapezoid. 

3.  No  two  sides  II :  Trapezium. 


Ex.  298.  If  it  is  required  to  prove  a  given  quadrilateral  a  rectangle, 
show  by  reference  to  §  243  that  the  logical  steps  are  to  prove  first  that  it 
is  a  parallelogram  ;  then  that  it  has  one  right  angle. 

Ex.  299.  If  a  given  quadrilateral  is  to  be  proved  a  square,  show  that 
the  only  additional  step  after  those  in  Ex.  298  is  to  prove  two  adjacent 
sides  equal. 

Ex.  300.  If  a  given  quadrilateral  is  to  be  proved  a  rhombus,  what 
are  the  three  steps  corresponding  to  those  given  in  Ex.  298  and  Ex.  299  ? 

Ex.  301.  Since  rectangles  and  rhomboids  are  parallelograms,  they 
possess  all  the  general  properties  of  parallelograms.  What  property  dif- 
ferentiates rectangles  from  rhomboids  (1)  by  definition  ?  (2)  by  proof  ? 
(See  Ex.  266  and  Ex.  267.) 

Ex.  302.  (a)  What  two  properties  that  have  been  proved  distinguish 
squares  from  other  rectangles  ?  (See  Ex.  277  and  Ex.  278.) 

(6)  What  two  properties  that  have  been  proved  distinguish  rhombuses 
from  other  rhomboids?  (See  Ex.  277  and  Ex.  278.) 

(c)  Show  that  the  two  properties  which  distinguish  squares  and  rhom- 
buses from  the  other  members  of  their  class  are  due  to  the  common  prop- 
erty possessed  by  squares  and  rhombuses  by  definition. 

Ex.  303.  The  mid-point  of  the  hypotenuse  of  a  right  triangle  is  equi- 
distant from  the  three  vertices. 

Ex.  304.  If  a  line  AB  of  given  length  is  moved  so  that  its  ends  always 
touch  the  sides  of  a  given  right  angle,  what  is  the  locus  of  the  mid-point 
of  AB? 


94 


PLANE   GEOMETRY 


PROPOSITION  XXXVIII.     THEOREM 

244.  If  three  or  more  parallel  lines  intercept  equal  seg- 
ments on  one  transversal,  they  intercept  equal  segments 
on  any  other  transversal. 


Given  II  lines  AG,  BH,  CJ,  DK,  etc.,  which  intercept  the 
equal  segments  AB,  BC,  CD,  etc.,  on  transversal  AF,  and  which 
intercept  segments  GH,  HJ,  JK,  etc.,  on  transversal  GL. 

To  prove     GH  =  HJ  =  JK,  etc. 


ARGUMENT 

1.  Draw  GM,  HN,  JR,  etc.  II  A  F. 

2.  Now  AGMB,  BHNC,  CJRD, 

etc.,  are  UJ. 

3.  .'.  GM  =  AB,  HN  =  BC,  JR 

=  CD,  etc. 

4.  And  AB  =  BC  =  CD,  etc. 

5.  .%  GM  =  HN  =  JR,  etc. 

6.  Again  GM,   HN,  JR,   etc., 

are  II  to  each  other. 

7.  ,-.  Zl  =  Z2  =  Z3,  etc. 

8.  And  Z4  =  Z5  =  Z6,  etc. 


REASONS 

1.  Parallel  line  post.     §  179. 

2.  By  def.  of  a  O.     §  220. 

3.  The   opposite   sides  of   a 

O  are  equal.     §  232. 

4.  By  hyp. 

5.  Ax.  1.     §  54,  1. 

6.  If  two  str.  lines  are  II  to  a 

third  str.  line,  they  are 
II  to  each  other.  §  180. 

7.  Corresponding  A  of  II  lines 

are  equal.     §  190. 

8.  If  two  A  have  their  sides 

II  right  to  right  and  left 
to  left,  they  are  equal. 
§  200,  a. 


BOOK   I 


9.    .-.      A  GHM  —  A  HJN  = 

AJKR,  etc. 


10.    .'.  GH  =  77,7  =  JK,  etc. 


Q.E.D. 


9.  Two  A  are  equal  if  a  side 
arid  the  two  adj.  A  of  one 
are  equal  respectively 
to  a  side  and  the  two  adj. 
A  of  the  other.  §  105. 
10.  Homol.  parts  of  equal 
figures  are  equal.  §  110. 

245.  Question.     Are  the  segments  that  the  parallels  intercept  on  one 
transversal  equal  to  the  segments  that  they  intercept  on  another  trans- 
versal ?    Illustrate. 

246.  Cor.  I.     The  line  bisecting  one  of  the  non-paral- 
lel   sides    of   a    trapezoid    and    parallel    to    the    bases 
bisects  the  other  of  the  non-parallel  sides  also. 

247.  Cor.    II.     The        line 


joining    the    mid-points  of 

the  non-parallel  sides  of  a  Ex 

trapezoid  is  (a}    parallel  to 

the  bases;  and  (b}  equal  to       A         H  D 

one  half  their  sum. 

HINT,     (a)  Prove  EF  II  7?  O  and  AD  by  the  indirect  method.    (6)  Draw 
GH  II  CD.    Prove  AH  =  GB  ;    then  prove  EF  =  GC  =\(GC  +  HD}  = 


248.  Cor.  m.     The  line  bisect- 
ing one  side  of  a  triangle  and 
parallel  to  another  side  bisects 
tlie  third  side. 

249.  Cor.  IV.    The 

line     joining     the     F^- 

mid-points    of   two  Xx\         / 

sides  of  a   triangle  ^c~ ~^/? 

is    parallel    to   the 

third  side  and  equal  to  one  half  the  third  side. 
HINT.  •   Draw  CF  II  BA.     Prove  CF=AE  =  EB. 


96  PLANE    GEOMETRY 


PROPOSITION  XXXIX.     PROBLEM 

250.   To  divide  a  given  straight  line  into  any  number 
of  equal  parts. 


,'X 

K 

>< 

P'-S 

R    -'A''' 

>x     ,! 

\ 

xx                               x 

\ 

.''' 

\ 

I                                   H 

B 

A 

Given   straight  line  AB. 

To  divide   AB  into  n  equal  parts. 

I.   Construction 

1.  Draw  the  unlimited  line  AX. 

2.  Take  any  convenient  segment,  as  AR,  and,  beginning  at  A, 
lay  it  off  n  times  on  AX. 

3.  Connect  the  nth  point  of  division,  as  JT,  with  B. 

4.  Through  the  preceding  point  of  division,  as  P,  draw  a  line 
PH  II  KB.     §  188. 

5.  Then  HB  is  one  ?ith  of  AB. 

6.  .*.  HB,  if  laid  off  successively  on  AB,  will  divide  AB  into 
n  equal  parts. 

II.  The  proof  and  discussion  are  left  as  an  exercise  for  the 
student. 

Ex.  305.  Divide  a  straight  line  into  7  equal  parts. 

Ex.  306.  Construct  an  equilateral  triangle,  having  given  the  perim- 
eter. 

Ex.  307.  Construct  a  square,  having  given  the  perimeter. 


251.  Def.  The  line  joining  the  mid-points  of  the  non- 
parallel  sides  of  a  trapezoid  is  called  the  median  of  the  trape- 
zoid.  

Ex.  308.     Show,  by  generalizing,  that  Cor.  Ill,  Prop.  XXXVIII,  may 

be  obtained  from  Cor.  I  and  Cor.  IV  from  Cor.  II. 


BOOK  I  97 

Ex.  309.  The  lines  joining  the  mid-points  of  the  sides  of  a  quadri- 
latefal  taken  in  order  form  a  parallelogram. 

Ex.  310.  What  additional  statement  can  you  make  if  the  quadri- 
lateral in  Ex.  309  is  an  isosceles  trapezoid  ?  a  rectangle  ?  a  rhombus  ? 
a  square  ? 

Ex.  311.  Lines  drawn  from  the  mid-point  of  the  base  of  an  isosceles 
triangle  to  the  mid-points  of  its  equal  sides  form  a  rhombus  or  a  square. 
When  is  the  figure  a  rhombus  ?  when  a  square  ? 

Ex.  312.  The  mid-points  of  the  sides  of  a  quadrilateral  and  the  mid- 
points of  its  two  diagonals  are  the  vertices  of  three  parallelograms  whose 
diagonals  are  concurrent. 

Ex.  313.     What  is  the  perimeter  of  each  parallelogram  in  Ex.  312  ? 

Ex.  314.     Construct  a  triangle,  given  the  mid-points  of  its  sides. 

Ex.  315.  Through  a  given  point  within  an  angle  construct  a  line, 
limited  by  the  two  sides  of  the  angle,  and  bisected  at  the  given  point. 

Ex.  316.  Every  diagonal  of  a  parallelogram  is  trisected  by  the  lines 
joining  the  other  two  vertices  with  the  mid-points  of  the  opposite  sides. 

Ex.  317.  If  a  triangle  inscribed  in  another  triangle  has  its  sides 
parallel  respectively  to  the  sides  of  the  latter,  its  vertices  are  the  mid- 
points of  the  sides  of  the  latter. 

Ex.  318.  If  the  lower  base  KT  of  trapezoid  ESTK  is  double  the 
upper  base  RS,  and  the  diagonals  intersect  at  0,  prove  OK  double  OS, 
and  Or  double  OH. 

Ex.  319.  Construct  a  trapezoid,  given  the  two  bases,  one  diagonal, 
and  one  of  the  non-parallel  sides. 

In  the  two  following  exercises  prove  the  properties  which  require  proof, 
state  those  which  follow  by  definition,  and  those  which  have  been  proved 
in  the  text  : 

Ex.  320.     Properties  possessed  by  all  trapezoids : 

(a)  Two  sides  of  a  trapezoid  are  parallel. 

(6)  The  two  angles  adjacent  to  either  of  the  non-parallel  sides  are 
supplementary. 

(c)  The  median  of  a  trapezoid  is  parallel  to  the  bases  and  equal  to  one 
half  their  sum. 

Ex.  321.     In  an  isosceles  trapezoid : 
(a)  The  two  non-parallel  sides  are  equal. 

(6)  The  angles  at  each  base  are  equal  and  the  opposite  angles  are 
supplementary. 

(c)  The  diagonals  are  equal. 


98  PLANE   GEOMETRY 

PROPOSITION  XL.     THEOREM 

252.   The  two  perpendiculars  to  the  sides  of  an  angle 
from  any  point  in  its  bisector  are  equal. 

,C 


B  DA 

Given   /.  ABC;  P  any  point  in  BR,  the  bisector  of  Z.ABG\  PD 
and  PE,  the  Js  from  P  to  BA  and  BC  respectively. 
To  prove    PD  =  PE. 

ARGUMENT  ONLY 

1.  In  rt.  A  DBP  and  PBE,  PB  =  PB. 

2.  Z  DBP  =  /.  PBE. 

3.  .-.  A  DBP  =  A  PBE. 

4.  .'.  PD  —PE.  Q.E.D. 

253.   Prop.  XL  may  be  stated  as  follows : 
Every  point  in  the  bisector  of  an  angle  is  equidistant  from  the 
sides  of  the  angle. 

Ex.  322.  Find  a  point  in  one  side  of  a  triangle  which  is  equidistant 
from  the  other  two  sides  of  the  triangle. 

Ex.  323.  Find  a  point  equidistant  from  two  given  intersecting  lines 
and  also  at  a  given  distance  from  a  fixed  third  line. 

Ex.  324.  Find  a  point  equidistant  from  two  given  intersecting  lines 
and  also  equidistant  from  two  given  parallel  lines. 

Ex.  325.     Find  a  point  equidistant  from  the  four  sides  of  a  rhombus. 

Ex.  326.     The  two  altitudes  of  a  rhombus  are  equal.     Prove. 

Ex.  327.  Construct  the  locus  of  the  center  of  a  circle  of  given  radius, 
which  rolls  so  that  it  always  touches  the  sides  of  a  given  angle.  Do  not 
prove. 


BOOK  I 


99 


PROPOSITION  XLI.     THEOREM 
(Opposite  of  Prop.  XL) 

254.   The  two  perpendiculars  to  the  sides  of  an  angle 
from  any  point  not  in  its  bisector  are  unequal. 

C 


B  G 

Given   Z.ABG\  P  any  point  not  in  BR,  the  bisector  of  Z.ABC\ 
PD  and  PE,  Js  from  P  to  BA  and  BG  respectively. 

To  prove    PD  =£  PE. 

• 
OUTLINE  or  PROOF 

Draw  FG  J_  BA  ;  draw  PG.     Then  FE  =  FG. 

Now  PF  +  FG  >  PG.      .'.PE>PG.      But  PG  >  PD. 

.'.  PE  >  PD. 

255.  Prop.  XLI  may  be  stated  as  follows : 

Every  point  not  in  the  bisector  of  an  angle  is  not  equidistant 
from  the  sides  of  the  angle. 

256.  Cor.  I.    (Converse  of  Prop.  XL).      Every  point  equi- 
distant from  the  sides  of  an  angle  li^es  in  tJie  bisector  of 
the  angle. 

HINT.     Prove  directly,  using  the  figure  for  §  252,  or  apply  §  137. 

257.  Cor.  II.     The  bisector  of  an  angle  is  the  locus  of 
all  points  equidistant  from,  the  sides  of  the  angle. 


Ex.  328.     What  is  the  locus  of  all  points  that  are  equidistant  from 
a  pair  of  intersecting  lines  ? 


100 


PLANE   GEOMETRY 


CONCURRENT   LINE   THEOREMS 
PROPOSITION  XLIT.     THEOREM 

258.  The  bisectors  of  the  angles  of  a  triangle  are  con- 
current in  a  point  which  is  equidistant  from  the  three 
sides  of  the  triangle. 


Given   A  ABC  with  AF,  BE,  CD  the  bisectors  of  A  A,  B,  and 
G  respectively. 

To  prove:    (a)    AF,  BE,  CD  concurrent  in  some  point  as  O; 
(b)  the  point  0  equidistant  from  AB,  BC,  and  CA. 


ARGUMENT 

1.   BE  and  CD  will  intersect  at 
some  point  as  0. 


2.  Draw  OL,  OH,  and    OG,  J§ 

from  O  to  AB,  BC,  and  CA 
respectively. 

3.  v  0  is  in  BE,  OL  *=  OH. 


4.  v  0  is  in  CD,  OG  =  OH. 

5.  .'.  OL  =  OG. 


REASONS 

1.  If  two  str.  lines  are  cut  by 

a  transversal  making  the 
sum  of  the  two  int.  A  on 
the  same  side  of  the 
transversal  not  equal  to 
2  rt.  A,  the  lines  are 
not  II.  §  194. 

2.  From  a  point  outside  a  line 

there  exists  one  and  only 
one  J_  to  the  line.  §  155. 

3.  The  two  J§  to  the  sides  of 

an  /-  from  any  point  in 
its  bisector  are  equal. 
§252. 

4.  Same  reason  as  3. 

5.  Things  equal  to  the  same 

thing  are  equal  to  each 
other.  §  54,  1. 


BOOK  I 


ARGUMENT 


6.    /.  AF,  the  bisector  of  Z.CAB, 
passes  through  0. 


7.  .•.  AF,  BE,  and  CD  are  con- 

current in  0. 

8.  Also  0  is  equidistant  from 

AB,  BC,  and  CA.       Q.E.D. 


REASONS 


6.  Every     point     equidistant 

from  the  sides  of  an  Z. 
lies  in  the  bisector  of 
the  Z.  §  256. 

7.  By  def  .  of  concurrent  lines. 

§  196. 

8.  By  proof,  OL  =  OH  =  OG. 


259.  Cor.  The  point  of  intersection  of  the  bisectors  of 
the  three  angles  of  a  triangle  is  the  locus  of  all  points 
equidistant  from  the  three  sides  of  the  triangle. 


Ex.  329.  Is  it  always  possible  to  find  a  point  equidistant  from  three 
given  straight  lines  ?  from  four  given  straight  lines  ? 

Ex.  330.  Find  a  point  such  that  the  perpendiculars  from  it  to  three 
sides  of  a  quadrilateral  shall  be  equal.  (Give  geometric  construction.) 

Ex.  331.  Prove  that  if  the  sides  AB  and  AC  of  a  triangle  ABC  are 
prolonged  to  E  and  F,  respectively,  the  bisectors  of  the  three  angles  BAC, 
EBC,  and  BCF  all  pass  through  a  point  which  is  equally  distant  from  the 
three  lines  AE,  AF,  and  BC.  Is  any  other  point  in  the  bisector  of  the 
angle  BAC  equally  distant  from  these  three  lines?  Give  reason  for 
your  answer. 

Ex.  332.  Through  a  given  point  P  draw  a  straight  line  such  that 
perpendiculars  to  it  from  two  fixed  points  Q  and  E  shall  cut  off  on  it 
equal  segments  from  P.  (HINT.  See  §  246.) 

Ex.  333.  Construct  the  locus  of  the  center  of  a  circle  of  given  radius, 
which  rolls  so  that  it  always  touches  the  sides  of  a  given  triangle.  Do  not 
prove. 

Ex.  334.  Find  the  locus  of  a  point  in  one  side  of  a  parallelogram  and 
equidistant  from  two  other  sides.  In  what  parallelograms  is  this  locus  a 
vertex  of  the  parallelogram  ? 

Ex.  335.  Find  the  locus  of  a  point  in  one  side  of  a  parallelogram  and 
equidistant  from  two  of  the  vertices  of  the  parallelogram.  In  what 
class  of  parallelograms  is  this  locus  a  vertex  of  the  parallelogram  ? 

Ex.  336.  Construct  the  locus  of  the  center  of  a  circle  of  given  radius 
which  rolls  so  that  it  constantly  touches  a  given  circumference.  Do  not 
prove, 


.PLANE   GEOMETRY 


PROPOSITION  XLIII.     THEOREM 

260.  The  perpendicular  bisectors  of  the  sides  of  a  tri- 
angle are  concurrent  in  a  point  which  is  equidistant  from 
the  three  vertices  of  the  triangle. 


1. 


Given  A  ABC  with  FG,  HK,  ED,  the  J_  bisectors  of  AB,  BC,  CA. 
To  prove :  (a)  FG,  HK,  ED  concurrent  in  some  point  as  0 ; 
(6)  the  point  0  equidistant  from  A,  B,  and  C. 

REASONS 

1.   Two  lines  J_  respectively  to 
two     intersecting     lines 
also  intersect.     §  195. 
Draw  OA,  OB,  and  OC. 
v  0  is  in  FG,  the  _L  bisector 
of    AB,    OB  =  OA ;    and 
v  0  is  in  DE,  the  _L  bi- 


ARGUMENT 

FG  and  ED  will  intersect  at 
some  point  as  0. 


G. 


sector  of  CA,  OC  =  OA. 
OB  =  OC. 

HK,   the   J_  bisector    of 
BC,  passes  through  0. 


FG,  HK,  and  ED  are  con- 
current in  0. 

1.    Also  O  is  equidistant  from 
A,  B,  and  (7,  Q.E.D. 


2.  Str.  line  post.  I.     §  54,  15. 

3.  Every  point  in  the  _L  bi- 

sector of  a  line  is  equi- 
distant from  the  ends  of 
that  line.     §  134. 
4    Ax.  1.     §  54,  1. 

5.  Every    point     equidistant 

from  the  ends  of  a  line, 
lies  in  the  J_  bisector  of 
that  line.  §  139. 

6.  By  def .  of  concurrent  linest 

§  196. 

7.  By  proof,  OA  =  OB  =  OC. 


BOOK  I 


103 


261.  Cor.     The  point  of  intersection  of  the  perpendic- 
ular bisectors  of  the  three  sides  of  a  triangle  is  the  locus 
of  all  points  equidistant  from  the  three  vertices  of  the 
triangle.  

Ex.  337.  Is  it  always  possible  to  find  a  point  equidistant  from  three 
given  points  ?  from  four  given  points  ? 

Ex.  338.  Construct  the  perpendicular  bisectors  of  two  sides  of  an 
acute  triangle,  and  then  construct  a  circle  whose  circumference  shall  pass 
through  the  vertices  of  the  triangle. 

Ex.  339.  Construct  a  circle  whose  circumference  shall  pass  through 
the  vertices  of  a  right  triangle. 

Ex.  340.  Construct  a  circle  whose  circumference  shall  pass  through 
the  vertices  of  an  obtuse  triangle. 

PROPOSITION   XLIV.     THEOREM 

262.  The  altitudes  of  a  triangle  are  concurrent. 

K  B  L 


Given   A  ABC  with  its  altitudes  AD,  BE,  and  CF. 
To  prove    AD,  BE,  and  CF  concurrent. 

OUTLINE  OP  PROOF 

Through  the  vertices  A,  B,  and  C,  of  triangle  ABC,  draw 
lines  ||  BC,  AC,  and  AB,  respectively.  Then  prove,  by  means 
of  UJ,  that  AD,  BE,  and  CF  are  the  _L  bisectors,  respectively,  of 
the  sides  of  the  auxiliary  A  HKL.  Then,  by  Prop.  XLIII,  AD, 
BE,  and  CF  are  concurrent.  Q.E.D. 


104  PLANE   GEOMETRY 

PROPOSITION   XLV.     THEOREM 

263.   Any  two  medians   of  a   triangle  intersect   each 
other  in  a  trisection  point  of  each. 

B 


Given   A  ABC  with  AD  and  CE  any  two  of  its  medians. 
To  prove   that  AD  and  CE  intersect  in  a  point  O  such  that 
OD  =  1  AD  and  OE  =  1  CE. 

OUTLINE  OF  PROOF 

1.  AD  and  CE  will  intersect  at  some  point  as  0.     §  194. 

2.  Let  R  and  5  be  the  mid-points  of  AO  and  CO  respectively. 

3.  Quadrilateral  REDS  is  a  O. 

4.  .-.  ^ff  =  #0  =  OD  and  CS  =  SO  =  OE. 

5.  That  is,  OD  =  J-  .4Z)  and  OE*=\  CE.  Q.E.D. 

264.  Cor.     J7ie   three  medians  of  a  triangle  are  con- 
current. 

265.  Def.     The  point  of   intersection  of  the  medians  of  a 
triangle  is  called  the  median  center  of  the  triangle.     It  is  also 
called  the  centroid  of  the  triangle.     This  point  is  the  center 
of  mass  or  center  of  gravity  of  the  triangle. 


Ex.  341.  Draw  a  triangle  whose  altitudes  will  intersect  on  one  of 
its  sides,  and  repeat  the  proof  for  Prop.  XLIV. 

Ex.  342.  Draw  a  triangle  whose  altitudes  will  intersect  outside  of 
the  triangle,  and  repeat  the  proof  for  Prop.  XLIV. 

Ex.  343.  Prove  Prop.  XLV  by  prolonging  OD  its  own  length  and 
drawing  lines  to  B  and  C  from  the  end  of  the  prolongation. 

Ex.  344.  Construct  a  triangle,  given  two  of  its  medians  and  the 
angle  between  them. 


BOOK  I 


105 


PROPOSITION  XLVI.     THEOREM 

266.   If  one  acute  angle  of  a  right  triangle  is  double 
the  other,  the  hypotenuse  is  double  the  shorter  side . 

B 


v 

1\ 


A  C 

Given    rt.  A  ABC  with  Z  A  double  Z  B. 
To  prove    AB  =  2  AC. 

ARGUMENT  ONLY 

1.  Draw'C%  making  Zl  = 

2.  Z  ^  =  60°. 

3.  .-.Zl=60°. 

5.  .\AE  =  AC=EC. 

6.  In  A  EBC,  Z  2  =  30°. 

7.  Z  z?  =  30°. 
8. 


Z  A. 


.\EB-EC. 


9. 
10. 


=  AE=  AC. 


Q.E.D. 


267.  Prop.  XLVI  is  sometimes  stated :    In  a  thirty-sixty  de- 
gree right  triaiujle,  the  hypotenuse  is  double  the  shorter  side. 

268.  Historical  Note.     Such  a  triangle  (a  30°-60°  right  triangle)  is 
spoken  of  by  Plato  as  "the  most  beautiful  right-angled  scalene  triangle/' 


Ex.  345.     Prove  Prop.  XLVI  by  drawing  CE  =  CA. 
Ex.  346.     Prove  Prop.  XLVI  by  drawing  lines  through  the  ends  of 
the  hypotenuse  parallel  to  the  other  two  sides,  thus  forming  a  rectangle. 


106 


PLANE   GEOMETRY 


CONSTRUCTION   OF   TRIANGLES 

269.  A  triangle  is  determined,  in  general,  when  three  parts  are 
given,  provided  that  at  least  one  of  the  given  parts  is  a  line. 

The  three  sides  and  the  three  angles  of  a  triangle  are  called 
its  parts ;  but  there  are  also  many  indirect  parts ;  as  the  three 
medians,  the  three  altitudes,  the  three  bisectors,  and  the  parts 
into  which  both  the  sides  and  the  angles  are  divided  by  these 
lines. 

A  A 


I 


B     C 


270.  The  notation  given  in  the  annexed  figures  may  be 
used  for  brevity : 

A,  B,  C,        the  angles  of  the  triangle  ;  in  a  right  triangle,  angle 

C  is  the  right  angle, 
a,  b}  c,        the  sides  of  the  triangle;  in  a  right  triangle,  c  is  the 

hypotenuse. 

ma)  mb)  mc,  the  medians  to  o,  6,  and  c  respectively. 
ha,  7ib,  7ic,     the  altitudes  to  a,  b,  and  c  respectively. 
tn>  tb>  to       the  bisectors  of  A,  B,  and  C  respectively. 
1M  du,  the  segments  of  a  made  by  the  altitude  to  a. 

say  ra,  the  segments  of  a  made  by  the  bisector  of  angle  A. 

271.  The  student  should  review  the  chief  cases  of  construc- 
tion   of   triangles  already  given:    viz.     a,  B,  c;  A,b,C',  a,  6,  c; 
right  triangles  :    a,  b ;  6,  c ;  b,  B ;  b,  A  ;  c>A;.  review  also  §  152. 


Ex.  347.     State  in  words  the  first  eight  cases  given  in  §  271. 


BOOK  I  107 

PROPOSITION  XLVII.     PROBLEM 

272.  To  construct  a  triangle  having  two  of  its  sides 
equal  respectively  to  two  given  lines,  and  the  angle  oppo- 
site one  of  these  lines  equal  to  a  given  angle. 


a 


B  /A     X 

FIG.  1. 

Given  lines  a  and  b,  and  /-  B. 
To  construct    A  ABC. 

I.    Construction 

1.  Draw  any  line,  as  BX. 

2.  At  any  point  in  BX,  as  B,  construct  Z  XBY=  to  the  given 
£  B.     §  125. 

3.  On  B  Y  lay  off  BC  =  a. 

4.  With  C  as  center  and  with  b  as  radius,  describe  an  arc 
cutting  BX  at  A. 

5.  A  ABC  is  the  required  A. 

II.   Tfie  proof  is  left  as  an  exercise  for  the  student. 

III.    Discussion 

(1)  b  may  be  greater  than  a ; 

(2)  b  may  equal  a ; 

(3)  b  may  be  less  than  a. 

(1)  If  b  >  a,  there  will  be  one  solution,  i.e.  one  A  and  only 
one  can  be  constructed  which  shall  contain  the  .given  parts. 
This  case  is  shown  in  Fig.  1. 

(2)  If  b  =  a,  the  A  will  be  isosceles.     The  construction  will 
be  the  same  as  for  case  (1). 


108  PLANE   GEOMETRY 


(3)  If  b  <  a.  Make  the  construction  as  for  case  (1).  If  b  > 
the  iL  from  G  to  5-3T,  there  will  be  two  solutions,  since  A  AB<7 
and  1)56',  Fig.  2,  both  contain  the  required  parts.  If  b  equals 
the  _L  from  C  to  BX  there  will  be  o«e  solution.  The  A  will  be  a 
rt.  A.  If  b  <  the  _L  from  C  to  BX,  there  will  be  no  solution. 

In  the  cases  thus  far  considered,  the  given  Z  was  acute. 
The  discussion  of  the  cases  in  which  the  given  Z  is  a  rt.  Z. 
and  in  which  it  is  an  obtuse  Z  is  left  to  the  student. 

273.  Question.     Why  is  (1)  the  only  case  possible  when  the  given 
angle  is  either  right  or  obtuse  ? 

274.  The  following  exercises  are  given  to  A 
illustrate  analysis  of  problems  and  to  show  the 

use  of  auxiliary  triangles  in  constructions. 


M      H    B 


Ex.  348.     Construct  a  triangle,  given  a,  ha,  ma-  FIG.  1. 

ANALYSIS.  Imagine  the  problem  solved  as  in 
Fig.  1,  and  mark  the  given  parts  with  heavy  lines. 
The  triangle  AHM  is  determined  and  may  be  made 
the  basis  of  the  construction. 


Ex.  349.     Construct  a  triangle,  given  6,  raa,  mc 

ANALYSIS.     From  Fig.  2  it  will  be  seen 'that  tri-     v  -" 

angle  AGO  may  be  constructed.     Its  three  sides  are 
known,  since  AO  =  f  ma  and  CO  =  f  mc. A 

Ex.  350.    Construct  a  triangle,  given  a,  ha,hc. 

ANALYSIS.    In  Fig.  3,  right  triangle  CHB  is 
determined.      The  locus  of  vertex  A  is  a  line 
parallel  to   CB,  so  that  the  distance  between    r 
it  and  CB  is  equal  to  ha,  FIG,  3. 


BOOK  I  109 


Ex.  351.     Construct  a  triangle,  given  &  ,c,  raa. 

ANALYSIS.  Triangle  ABK,  Fig.  4,  is  deter- 
mined  by  three  sides,  6,  c,  and  2wa.  Since 
ABKC  is  a  parallelogram,  ylTf  bisects  (75. 

Ex.  352.  Construct  a  triangle,  given  a,  w«, 
and  the  angle  between  ma  and  a. 

ANALYSIS.  Triangle  ^3f  0  is  determined,  as 
shown  in  Fig.  5. 

A 


M 


y\    v 


C  M  B  T         S    «         N 

FIG.  5.  FIG-  C. 

Ex.  353.     Construct  a  trapezoid,  given  its  four  sides. 

ANALYSIS.     Triangle  R8T,   Fig.   6,    is   determined. 

Ex.  354.     Construct  a  parallelogram,  given  £ 


the  perimeter,  one  base  angle,  and  the  altitude.  ^T~  ~~7 

CONSTRUCTION.     The  two  parallels,  CH  and  s'     f\  I    / 

AE,  Fig.  7,  may  be  drawn  so  that  the  distance  A         B  E 

between  them  equals  the  altitude  ;  at  any  point  FlG-  7- 

B  construct  angle  EBC  equal  to  the  given  R M 

base  angle ;  draw  CA,  bisecting  angle  FOB  ;  /\  \ 

measure  AE  equal  to  half  the  given  perimeter  ;  /      \  \ 

complete  parallelogram  CHEB.  75  N 

Ex.  355.     Construct  a  trapezoid,  given  the  Fl«-  8- 

two  non-parallel  sides  and  the  difference  between  the  bases.     (See  Fig.  8.) 

Construct  a  triangle,  given  : 

Ex.  356.    A,  ha,  ta.  Ex.  360.     B,hab. 

Ex.  357.     ha)  la,  da-  Ex.  361.     sa,  tn,  ha. 

Ex.  358.    ha,  ma,  la.  Ex.  362.     a,  r»6,  C. 

Ex.  359.     a,  tb,  C.  Ex.  363.     a,  «&,  B. 

Construct  an  isosceles  trapezoid,  given : 

Ex.  364.   The  two  bases  and  the  altitude. 

Ex.  365.    One  base,  the  altitude,  and  a  diagonal. 

Ex.  366.   A  base,  a  diagonal,  and  the  angle  between  them. 


110  PLANE   GEOMETRY 

DIRECTIONS   FOR   THE   SOLUTION    OF    EXERCISES 

275.  I.  Make  the  figures  clear,  neat,  accurate,  and  general 
in  character. 

II.  Fix  firmly  in  mind  hypothesis  and  conclusion  with  ref- 
erence to  the  given  figure. 

III.  Recall  fundamental  propositions  related  to  the  propo- 
sition in  question. 

IV.  If  you  can  find  no  theorem  which  helps  you,  contradict 
the  conclusion  in  every  possible  way   (reductio  ad  absurdum) 
and  try  to  show  the  absurdity  of  the  contradiction. 

V.  Make  frequent  use  of  the  method  of  analysis,  which  con- 
sists in  assuming  the  proposition  proved,  seeing  what  results 
follow  until  a  known  truth  is  reached,  and  then  retracing  the 
steps  taken. 

VI.  If  it  is  required  to  find  a  point  which   fulfills  two 
conditions,   it  is  often  convenient  to  find  the  point  by  the 
Intersection  of  Loci.     By  finding  the  locus  of  a  point   which 
satisfies  each  condition  separately,  it  is  possible  to  find  the 
points  in  which  the  two  loci  intersect;  i.e.  the  points  which 
satisfy  both  conditions  at  the  same  time. 

VII.  See  §  152  and  exercises  following  §  274  for  method  of 
attacking  problems  of  construction. 

The  method  just  described  under  V  is  a  shifting  of  an 
uncertain  issue  to  a  certain  one.  It  is  sometimes  called  the 
Method  of  Successive  Substitutions.  It  may  be  illustrated  thus : 

1.  A  is  true  if  B  is  true.  3.    But  C  is  true. 

2.  B  is  true  if  C  is  true.  4.    .-.  A  is  true. 

This  is  also  called  the  Analytic  Method  of  proof.  The 
proofs  of  the  theorems  are  put  in  what  is  called  the  Synthetic 
form.  But  these  were  first  thought  through  analytically,  then 
rearranged  in  the  form  in  which  we  find  them. 


BOOK  I  111 

MISCELLANEOUS   EXERCISES 

Ex.  367.  The  perpendiculars  drawn  from  the  extremities  of  one 
side  of  a  triangle  to  the  median  upon  that  side  are  equal. 

Ex.  368.     Construct  an  angle  of  75°  ;  of  97  £°. 

Ex.  369.  Upon  a  given  line  find  a  point  such  that  perpendiculars 
from  it  to  the  sides  of  an  angle  shall  be  equal. 

Ex.  370.  Construct  a  triangle,  given  its  perimeter  and  two  of  its 
angles. 

Ex.  371.  Construct  a  parallelogram,  given  the  base,  one  base  angle, 
and  the  bisector  of  the  base  angle. 

Ex.  372.  Given  two  lines  that  would  meet  if  sufficiently  prolonged. 
Construct  the  bisector  of  their  angle,  without  prolonging  the  lines. 

Ex.  373.  Construct  a  triangle,  having  given  one  angle,  one  adjacent 
side,  and  the  difference  of  the  other  two  sides.  Case  1  :  The  side  oppo- 
site the  given  angle  less  than  the  other  unknown  side.  Case  2  :  The  side 
opposite  the  given  angle  greater  than  the  other  unknown  side. 

Ex.  374.  The  difference  between  two  adjacent  angles  of  a  parallelo- 
gram is  90°  ;  find  all  the  angles. 

Ex.  375.  A  straight  railway  passes  2  miles  from  a  certain  town.  A 
place  is  described  as  4  miles  from  the  town  and  1  mile  from  the  railway. 
Represent  the  town  by  a  point  and  find  by  construction  how  many  places 
answer  the  description. 

Ex.  376.  Describe  a  circle  through  two  given  points  which  lie  out- 
side a  given  line,  the  center  of  the  circle  to  be  in  that  line.  Show  when 
no  solution  is  possible. 

Ex.  377.  Construct  a  right  triangle,  given  the  hypotenuse  and  the 
difference  of  the  other  two  sides. 

Ex.  378.  If  two  sides  of  a  triangle  are  unequal,  the  median  through 
their  intersection  makes  the  greater  angle  with  the  lesser  side. 

Ex.  379.  Two  trapezoids  are  equal  if  their  sides  taken  in  order  are 
equal,  each  to  each. 

Ex.  380.  Construct  a  right  triangle,  having  given  its  perimeter  and 
an  acute  angle. 

Ex.  381.  Draw  a  line  such  that  its  segment  intercepted  between 
two  given  indefinite  lines  shall  be  equal  and  parallel  to  a  given  finite  line. 

Ex.  382.  One  angle  of  a  parallelogram  is  given  in  position  and  the 
point  of  intersection  of  the  diagonals  is  given  ;  construct  the  parallelo- 
gram. 


112 


PLANE   GEOMETRY 


Ex.  383.  Construct  a  triangle,  given  two  sides  and  the  median  to  the 
third  side. 

Ex.  384.  If  from-  any  point  within  a  triangle  lines  are  drawn  to  the 
three  vertices  of  the  triangle,  the  sum  of  these  lines  is  less  than  the  sum  of 
the  sides  of  the  triangle,  and  greater  than  half  their  sum. 

Ex,  385.  Repeat  the  proof 
of  Prop.  XIX  for  two  cases  at 
once,  using  Figs.  1  and  2. 

Ex.  386.  If  the  angle  at  the 
vertex  of  an  isosceles  triangle 
is  four  times  each  base  angle, 
the  perpendicular  to  the  base 
at  one  end  of  the  base  forms  FIG.  1.  FIG.  2. 

with  one  side  of  the   triangle,  and  the  prolongation  of  the  other  side 
through  the  vertex,  an  equilateral  triangle. 

Ex.  387.  The  bisector  of  the  angle  C  of  a  triangle  ABC  meets  AB 
in  Z>,  and  DE  is  drawn  parallel  to  AC  meeting  BC  in  E  and  the  bisector 
of  the  exterior  angle  at  C  in  F.  Prove  DE  =  EF. 

Ex.  388.  Define  a  locus.  Find  the  locus  of  the  mid-points  of  all 
the  lines  drawn  from  a  given  point  to  a  given  line  not  passing  through 
the  point. 

Ex.  389.  Construct  an  isosceles  trapezoid,  given  the  bases  and  one 
angle. 

Ex.  390.  Construct  a  square,  given  the  sum  of  a  diagonal  and  one 
side. 

Ex.  391.  The  difference  of  the  distances  from  any  point  in  the  base 
prolonged  of  an  isosceles  triangle  to  the  equal  sides  of  the  triangle  is 
constant. 

Ex.  392.  Find  a  point  X  equidistant  from  two  intersecting  lines 
and  at  a  given  distance  from  a  given  point. 

Ex.  393.  When  two  lines  are  met  by  a  transversal,  the  difference  of 
two  corresponding  angles  is  equal  to  the  angle  between  the  two  lines. 

Construct  a  triangle,  given  : 

Ex.  394.     .4,  ha,  la-  Ex    398.  a,  m(l,  B. 

Ex.  395.     A,  ta,  sa.  Ex.   399.  wi,.,  7tc,  B. 

Ex.  396.     a,  hu,  la.  Ex.  400.  &,  c,  B  +  C. 

Ex.  397.     a,  b  +  c,  A.  Ex.  401.  A,  B,  b  +  c. 


BOOK  II 


THE   CIRCLE 

276.  Def.     A  circle  is  a  plane  closed  figure  whose  boundary 
is  a  curve  such  that  all  straight  lines  to  it  from  a  fixed  point 
within  are  equal. 

277.  Def.     The  curve  which  forms 
the  boundary  of  a  circle  is  called  the 
circumference. 


278.  Def.     The  fixed  point  within 
is  called  the  center,  and  a  line  joining 
the  center  to  any  point  on  the  circum- 
ference is  called  a  radius,  as  QR. 

279.  From  the  above  definitions  and  from   the   definition 
of  equal  figures,  §  18,  it  follows  that : 

(a)  All  radii  of  the  same  circle  are  equal. 

(6)   All  radii  of  equal  circles  are  equal. 

(c)   All  circles  having  equal  radii  are  equal. 

280.  Def.     Any  portion  of 

a  circumference  is  called  an  H 

arc,  as  DF,  FC,  etc. 

281.  Def.  A  chord  is  any 

straight  line  having  its  ex- 
tremities on  the  circumfer- 
ence, as  DF. 

282.  Def.      A  diameter  is  a 

chord  which  passes  through 
the  center,  as  BC. 

283.  Since  any  diameter  is  twice  a  radius,  it  follows  that 

All  diameters  of  a  circle  are  equal. 

113 


114 


PLANE   GEOMETRY 


PROPOSITION  I.     THEOREM 

284.   Every  diameter  of  a  circle  bisects  tlw  circumfer- 
ence and  the  circle. 

M 


Given   circle  AMBN  with  center  O,  and  AB,  any  diameter. 
To  prove:  (a)  that  AB  bisects  circumference 
(?>)  that  AB  bisects  circle  AMBN. 


ARGUMENT 

1.    Turn  figure  AMB  on  AB  as  an  axis  until 

it  falls  upon  the  plane  of  ANB. 
Arc  AMB  will  coincide  with  arc  ANB. 
.\  arc  AMB  =  arc  ANB ;    i.e.  AB  bisects 

circumference  AMBN. 
Also  figure  AMB  will  coincide  with  figure 

ANB. 
.-.figure    AMB  =  figure    ANB;     i.e.    AB 

bisects  circle  AMBN.  Q.E.D. 


REASONS 

1.  §  54,  14. 

2.  §  279,  «. 

3.  §  18. 

4.  §  279,  a. 

5.  §  18. 


Ex.  402.     A  semicircle  is  described  upon  each  of  the  diagonals  of  a 
rectangle  as  diameters.     Prove  the  semicircles  equal. 

Ex.  403.     Two  diameters  perpendicular  to  each  other  divide  a  circum- 
ference into  four  equal  arcs.     Prove  by  superposition. 

Ex.  404.     Construct  a  circle  which  shall  pass  through  two  given  points. 

Ex.  405.     Construct   a  circle  having  a  given  radius  r,  and  passing 
through  two  given  points  A  and  B. 


BOOK   II 


115 


285.  Def .     A  secant  of  a  cir- 
cle  is   a   straight   line   which 
cuts  the  circumference  in  two 
points,  but  is  not  terminated 
by  the  circumference,  as  MN. 

286.  Def.     A   straight   line 
is  tangent  to,  or  touches,  a  cir- 
cle if,  however  far  prolonged, 
it  meets  the  circumference  in 
but  one  point.     This  point  is 

called   the   point   of  tangency.     HK  is  tangent  to  circle    0   at 
point  T,  and  T  is  the  point  of  tangency. 

287.  Def.     A  sector  of  a  circle  is  a  plane  closed  figure  whose 
boundary  is  composed  of  two  radii  and  their  intercepted  arc, 
as  sector  SOR. 

288.  Def.    A  segment  of  a  circle 
is   a   plane    closed    figure   whose 
boundary  is   composed  of  an  arc 
and  the  chord  joining  its  extremi- 
ties, as  segment  DCE. 

289.  Def.     A  segment  which  is 
one  half  of  a  circle  is  called  a  semi- 
circle, as  segment  AMB. 

290.  Def.     An  arc  which  is  half 

of  a  circumference  is  called  a  semicircumference,  as  arc  AMB. 

291.  Def.     An   arc   greater   than    a    semicircumference    is 
called   a  major  arc,  as  arc  DME;  an  arc  less  than  a  semicir- 
cumference is  called  a  minor  arc,  as  arc  DCE. 

292.  Def.     A  central  angle,  or  angle  at  the  center,  is  ail  angle 
whose  vertex  is  at  the  center  of  a  circle  and  whose  sides  are 
radii. 

Ex.  406.     The  line  joining  the  centers  of  two  circles  is  0,  the  radii  are 
8  and  10,  respectively.     What  are  the  relative  positions  of  the  two  circles  ? 
Ex.  407.     A  circle  can  have  only  one  center. 


116 


PLANE   GEOMETRY 


PROPOSITION  II.     THEOREM 

293.  In  equal  circles,  or  in  the  same  circle,  if  two  cen- 
tral angles  are  equal,  tfay  intercept  equal  arcs  on  the  cir- 
cumference; conversely,  if  two  arcs  are  equal,  the  central 
angles  that  intercept  them  are  equal. 


I.     Given   equal  circles  ABM  and  CDN,  and  equal  central  A  0 
and  Q,  intercepting  arcs  AB  and  CD,  respectively. 
To  prove    A~B  =  CD. 


ARGUMENT 

1.  Place   circle  ABM  upon   circle  CDN  so 

that  center  0  shall  fall  upon  center  Q, 
and  OA  shall  be  collinear  with  QC. 

2.  A  will  fall  upon  <7. 

3.  OB  will  become  collinear  with  QD. 

4.  .-.  B  will  fall  upon  D. 

5.  .*.  AB  will  coincide  with  CD. 

6.  .'.  AB  =  CD. 


Q.E.D. 


RKASONS 
1.    §54,14. 


2.  §  279,  b. 

3.  By  hyp. 

4.  §  279,  b. 

5.  §279,6. 

6.  §18. 


II. 


Conversely : 

Given   equal  circles  ABM  and  CDN,  and  equal  arcs  AB  and  CD, 
intercepted  by  A  0  and  Q,  respectively. 
To  prove   Zo  -  ZQ. 


BOOK   II 


117 


ARGUMENT 

1.  Place  circle  ABM  upon  circle   CDN  so 

that  center  0  shall  fall  upon  center  Q. 

2.  Rotate   circle  AB  M  upon  0  as  a  pivot 

until  AB  falls  upon  its  equal  cbj  A 
upon  C,  B  upon  D. 

3.  O4  will  coincide  with  QC'and  OB  with  QD. 

4.  .-.  Z  0  =  Z  Q.  Q.E.D. 


REASONS 

1.  §  54,  14. 

2.  §54,14. 


3.  §17. 

4.  §18. 


294.  Cor.     In  equal  circles,  or  in  the  same  circle,  if 
two  central  angles  are  unequal,  the  greater  angle  inter- 
cepts the  greater  arc ;  conversely,  if  two  arcs  are  unequal, 
the  central  angle  that  intercepts  the  greater  arc  is  the 
greater.     (HINT.     Lay  off  the  smaller  central  angle  upon  the  greater.) 

295.  fcef.     A  fourth   part  of  a   circumference   is   called  a 
quadrant. 

From  Prop.  II  it  is  evident 
that  a  right  angle  at  the  center 
intercepts  a  quadrant  on  the  cir- 
cumference. Thus,  two  _L  diam- 
eters AB  and  CD  divide  the 
circumference  into  four  quadrants, 
AC,  CB,  BD,  and  DA. 

293.  Def.  A  degree  of  arc,  or  an 
arc  degree,  is  the  arc  intercepted 
by  a  central  angle  of  one  degree. 

297.  A  right  angle  contains  ninety  angle  degrees  (§  71) ; 
therefore,  since  equal  central  angles  intercept  equal  arcs  on  the 
circumference,  a  quadrant  contains  ninety  arc  degrees. 

Again,  four  right  angles  contain  360  angle  degrees,  and  four 
right  angles  at  the  center  of  a  circle  intercept  a  complete  cir- 
cumference ;  therefore,  a  circumference  contains  360  arc  degrees. 
Hence,  a  semicircumference  contains  180  arc  degrees. 

Ex.  408.  Divide  a  given  circumference  into  eight  equal  arcs  ;  sixteen, 
equal  arcs. 


118 


PLANE   GEOMETRY 


Ex.  409.  Divide  a  given  circumference  into  six  equal  arcs ;  three 
equal  arcs ;  twelve  equal  arcs. 

Ex.  410.  A  diameter  and  a  secant  perpendicular  to  it  divide  a  cir- 
cumference into  two  pairs  of  equal  arcs. 

Ex.  411.  Construct  a  circle  which  shall  pass  through  two  given  points 
A  and  B  and  shall  have  its  center  in  a  given  line  c. 

Ex.  412.  If  a  diameter  and  another  chord  are  drawn  from  a  point  in 
a  circumference,  the  arc  intercepted  by  the  angle  between  them  will  be 
bisected  by  a  diameter  drawn  parallel  to  the  chord. 

Ex.  413.  If  a  diameter  and  another  chord  are  drawn  from  a  point  in 
a  circumference,  the  diameter  which  bisects  their  intercepted  arc  will  be 
parallel  to  the  chord. 


PROPOSITION  III.     THEOREM 

298.  In  equal  circles,  or  in  the  same  circle,  if  two 
chords  are  equal,  they  subtend  equal  arcs;  conversely, 
if  two  arcs  are  equal,  the  chords  that  subtend  them  are 
equal. 


I.    Given  equal  circles  0  and  Q,  with  equal  chords  AB  and  CD. 
To  prove    AB  =  CD. 


1. 

2 

3. 

4. 
5. 

G. 


ARGUMENT 

Draw  radii  OA,  OB,  QC,  QD. 
In  A  OAB  and  QCD,  AB  =  CD. 
OA  =  QC  and  OB  =  QD. 
.».  A  OAB  =  A  QCD. 
.-.  ZO  =  ZQ. 

.-.  AB=  CD. 


Q.K.D. 


REASONS 

§  54,  15. 
By  hyp. 
§  279,  b. 
§116. 
§110. 
§  293,  I. 


BOOK  II 


119 


II.    Conversely : 

Given     equal  circles  O  and  Q,  and  equal  arcs  AB  and  CD. 

To  prove     chord  AB  =  chord  CD. 


ARGUMENT 

1.  Draw  radii  OA,  OB,  QC,  QD. 

2.  AB  =  CD. 

4.  OA  =  QC  and  OB  =  QD. 

5.  .'.  A  OAB  =  A  QCD. 

6.  .-.  chord  AB  =  chord  CD. 


Q.E.D. 


REASONS 
§  54,  15. 

2.  By  hyp. 

3.  §293,11. 

4.  §  279,  b. 

5.  §107. 

6.  §110. 


1. 


Ex.  414.  If  a  circumference  is  divided  into  any  number  of  equal  arcs, 
the  chords  joining  the  points  of  division  will  be  equal. 

Ex.  415.     A  parallelogram  inscribed  in  a  circle  is  a  rectangle. 

Ex.  416.  If  two  of  the  opposite  sides  of  an  inscribed  quadrilateral 
are  equal,  its  diagonals  are  equal. 

Ex.  417.     State  and  prove  the  converse  of  Ex.  416. 


299.  Def.    A  polygon  is  inscribed 
in  a  circle  if  all  its  vertices  are  on 
the  circumference.     Thus,  polygon 
ABCDE  is  an  inscribed  polygon. 

300.  Def.    If   a   polygon    is    in- 
scribed in  a  circle,  the  circle  is  said 
to  be  circumscribed  about  the  poly- 
gon. 


Ex.  418.  Inscribe  an  equilateral  hexagon  in  a  circle ;  an  equilateral 
triangle/ 

Ex.  419.    The  diagonals  of  an  inscribed  equilateral  pentagon  are  equal. 

Ex.  420.  If  the  extremities  of  any  two  intersecting  diameters  are 
joined,  an  inscribed  rectangle  will  be  formed.  Under  what  conditions 
will  tho  rectangle  be  a  square  ? 

Ex.  421.  State  the  theorems  which  may  be  used  in  proving  arcs  equal. 
State  the  theorems  which  may  be. used  in  proving  chords  equal. 


120 


PLANE   GEOMETRY 


PROPOSITION  IV.     THEOREM 

301.  In  equal  circles,  or  in  tlie  same  circle,  if  two  chords 
are  unequal,  the  greater  chord  subtends  tlie  greater  minor 
arc ;  conversely,  if  two  minor  arcs  are  unequal,  the  chord 
that  subtends  the  greater  arc  is  tlie  greater. 


I.  Given   circle  0,  with  chord  AB  >  chord  CD. 
To  prove    AB  >  CD. 

ARGUMENT  REASONS 

1.  Draw  radii  OA,  OB,  OC,  OD.  1.  §  54,  15. 

2.  In  A  OAB  and  OCD,  OA  =  OC,  OB  =  OD.  2.  §  279,  a. 

3.  Chord  AB  >  chord  CD.  3.  By  hyp. 

4.  .-.  Zl  >  Z2.  4.  §  173. 

5.  .-.  AB  >  CD.                                     Q.E.D.  5.  §  294. 

II.  Conversely: 

Given     circle  O,  with  AB  >  CD. 
To  prove     chord  AB  >  chord  CD. 

ARGUMENT  REASONS 

1.  Draw  radii  OA,  OB,  OC,  OD.  1.  §  54,  15. 

2.  In  A  OAB  and  OCD,  OA  =  OC,  OB  =  OD.  2.  §  279,  a. 

3.  AB  >  CD.  3.  By  hyp. 

4.  .-.  Zl  >  Z2.  4.  §294. 

5.  .-.  chord  AB  >  chord  OD.                Q.E.D.  5.  §  172. 

Ex.  422..   Prove  the  converse  of  Prop.  IV  by  the  indirect  method. 


BOOK   II 


121 


PROPOSITION  V.     THEOREM 

302.   The   diameter  perpendicular  to  a  chord  bisects 
w  chord  and  also  its  subtended  arcs. 


REASONS 

1. 

§ 

54,  15. 

2. 

§ 

279,  a. 

a 

§ 

94. 

4. 

§ 

212. 

5. 

§ 

212. 

6. 

§ 

75. 

7. 

§ 

293,  I. 

Given     chord  AB  and  diameter  CD  J_  AR  at  E. 
To  prove     AE  =  EB,  AC  =  CB,  and  AD  =  DB. 

ARGUMENT 

1.  Draw  radii  OA  and  OB. 

2.  In  A  OAB,  OA  =  OB. 

3.  .-.A  OAB  is  an  isosceles  A. 

4.  .-.  OE  bisects  AB,  and  AE  =  EB. 

5.  Also  OE  bisects  /.  BOA,  and  Zl  =  Z2. 

6.  .-.  ^AOD  =  ZDOB. 

1.   .-.  AC=  CB  and  AD  =  DB.  Q.E.D. 

303.  Cor.   I.    The  perpendicular   bisector   of  a    chord 
passes  through  the  center  of  the  circle. 

304.  Cor.  II.    The   locus  of  the   centers   of  all   circles 
ivhich  pass  through  two  given  points  is  the  perpendicu- 
lar bisector  of  the  line  which  joins  the  points. 

305.  Cor.  in.    Tlw  locus  of  the  mid-points  of  all  chords 
of  a  circle  parallel  to  a  given  line  is  the  diameter  per- 
pendicular to  the  line. 

Ex.  423.    If  the  diagonals  of  an  inscribed  quadrilateral  are  unequal,  its 
opposite  sides  are  unequal. 


122 


PLANE   GEOMETRY 


Ex.  424.  Through  a  given  point  within  a  circle  construct  a  chord 
which  shall  be  bisected  at  the  point. 

Ex.  425.  Given  a  line  fulfilling  any  two  of  the  five  following  condi- 
tions, prove  that  it  fulfills  the  remaining  three: 

1.  A  diameter. 

2.  A  perpendicular  to  a  chord. 

3.  A  bisector  of  a  chord. 

4.  A  bisector  of  the  major  arc  of  a  chord. 

5.  A  bisector  of  the  minor  arc  of  a  chord. 

Ex.  426.  Any  two  chords  of  a  circle  are  given  in  position  and  magni- 
tude ;  find  the  center  of  the  circle. 

Ex.  427.  The  line  passing  through  the  middle  points  of  two  parallel 
chords  passes  through  the  center  of  the  circle. 

Ex.  428.     Given  an  arc  of  a  circle,  find  the  center  of  the  circle. 

PROPOSITION  VI.     PROBLEM 
306.    To  bisect  a  given  arc. 

A 


Given     AB,  an  arc  of  any  circle. 
To  bisect     AB. 

The    construction,    proof,    and    discussion    are    left    as    an 
exercise  for  the  student. 


Ex.  429.  Construct  an  arc  of  45° ;  of  30°.  Construct  an  arc  of  30°, 
using  a  radius  twice  as  long  as  the  one  previously  used.  Are  these  two 
30°  arcs  equal  ? 

Ex.  430.  Distinguish  between  finding  the  "  mid-point  of  an  arc  '"and 
the  "  center  of  an  arc." 


BOOK  II 


123 


PROPOSITION  VII.     THEOREM 

307.  In  equal  circles,  or  in  the  same  circle,  if  two  chords 
are  equal,  they  are  equally  distant  from  the  center;  con- 
versely, if  two  chords  are  equally  distant  from  the  center, 
they  are  equal. 


B 


I.    Given   circle  0  with  chord  AB  =  chord  CD,  and  let  OE  and 
OF  be  the  distances  of  AB  and  CD  from  center  O,  respectively. 
To  prove    OE  =  OF. 

ARGUMENT 

1.  Draw  radii  OB  and  0(7. 

2.  E  and  F  are  the  mid-points  of  AB  and 

CD,  respectively. 

.'.  in  rt.  A  OEB  and  OCF,  EB  =  CF. 
OB  =  OC.      . 
.-.  A  OEB  =  A  OCF. 

.'.   OE  —  OF.  Q.E.D. 


REASONS 
§  54,  15. 
§  302. 


3.  §  54,  8  a. 

4.  §  279,  a. 

5.  §  211. 

6.  §  110. 


II.    Conversely : 

Given   circle  O  with  OE,  the  distance  of  chord  AB  from  center 
0,  equal  to  OF,  the  distance  of  chord  CD  from  center  O. 
To  prove    chord  AB  =  chord  CD. 
HINT.     Prove  A  OEB  =  A  OCF. 


Ex.  431.  If  perpendiculars  from  the  center  of  a  circle  to  the  sides  of 
an  inscribed  polygon  are  equal,  the  polygon  is  equilateral. 

Ex.  432.  If  through  any  point  in  a  diameter  two  chords  are  drawn 
making  equal  angles  with  the  diameter,  the  two  chords  are  equal. 


124 


PLANE   GEOMETRY 


PROPOSITION  VIII.     THEOREM 

308.  In  equal  circles,  or  in  the*  same  circle,  if  two 
chords  are  unequal,  the  greater  chord  is  at  the  less  dis- 
tance from  the  center. 


Given    circle  O  with  chord  AB  >  chord  CD,  and  let  GF  and 
OH  be  the  distances  of  AB  and  CD  from  center  0,  respectively. 
To  prove    OF  <  OH. 


ARGUMENT 

1.  From  A  draw  a  chord  AE,  equal  to  DC. 

2.  From  0  draw  OG  ±  AE. 

3.  Draw  FG. 

4.  AB  >  CD. 

5.  .-.  AB  >  AE. 

6.  F  and  G  are  the  mid-points  of  AB  and 

.4J?,  respectively. 
.'.  AF>  AG. 
.-.41  >Z2. 
Z  JFO  =  Z 
.-.  Z3  <Z4. 


7. 

8. 

9. 
10. 
11. 
12. 
13. 


0£  =  OJI. 
.'.  OF<  OH. 


Q.E.D. 


REASONS 

1.  §  54,  15. 

2.  §  155. 

3.  §  54,  15. 

4.  §  By  hyp. 

5.  §  309. 

6.  §  302. 


§  54,  8  b. 
§  156. 
§  64. 
§  54,  6. 
§  164. 

12.  §  307,  I. 

13.  §  309. 


8. 

9. 
10. 
11. 


309.  Note.  The. student  should  give  the  full  statement  of  the  sub- 
stitution made  ;  thus,  reason  5  above  should  be  :  "  Substituting  AE  for 
its  equal  <7Z>," 


BOOK  II  125 

PROPOSITION  IX.     THEOREM 
(Converse  of  Prop.  VIII) 

310.  In  equal  circles,  or  in  the  same  circle,  if  two  chords 
are  unequally  distant  from  the  center,  the  chord  at  the 
less  distance  is  the  greater. 


Given   circle  0  with  OF,  the  distance  of  chord  AB.  from  center 
0,  less  than  Off,  the  distance  of  chord  CD  from  center  0. 
To  prove    chord  AB  >  chord  CD. 
The  proof  is  left  as  an  exercise  for  the  student. 
HINT.     Begin  with  A  OGF. 

311.  Cor.   I.     A  diameter   is  greater   than  any  other 
chord. 

312.  Cor.  II.     The  locus  of  the  mid-points  of  all  chords 
of  a  circle  equal  to  a  given  chord  is  the  circumference  hav- 
ing the  same  center  as  the  given  circle,  and  having  for  ra- 
dius the  perpendicular  from  the  center  to  tfo  given  chord. 


Ex.  433.     Prove  Prop.  IX  by  the  indirect  method. 

Ex.  434.  Through  a  given  point  within  a  circle  construct  the  mini- 
mum chord. 

Ex.  435.  If  two  chords  are  drawn  from  one  extremity  of  a  diameter, 
making  unequal  angles  with  it,  the  chords  are  unequal. 

Ex.  436.  The  perpendicular  from  the  center  of  a  circle  to  a  side  of 
an  inscribed  equilateral  triangle  is  less  than  the  perpendicular  from  the 
center  of  the  circle  to  a  side  of  an  inscribed  square.  (See  §  308.) 


126 


PLANE   GEOMETRY 


PROPOSITION  X.     THEOREM 

313.  A  tangent  to  a  circle  is  perpendicular  to  the  radius 
drawn  to  the  point  of  tangency. 


A         M  T  B 

Given  line  AB,  tangent  to  circle  0  at  T,  and  OT,  a  radius  drawn 
to  the  point  of  tangency. 
To  prove    AB  J_  OT. 

ARGUMENT 

1.  Let  M  be  any  point  on  AB  other  than  T\ 

then  M  is  outside  the  circumference. 

2.  Draw  OM,  intersecting  the  circumference 

at  S. 

3.  OS  <  OM. 

4.  OS  =  OT. 

5.  .'.  OT  <  OM. 

6.  .-.  or  is  the  shortest  line  that  can  be 

drawn  from  0  to  AB. 

7.  .'.  OT±.AB  ;   i.e.  AB±.  OT.  Q.E.D. 

314.  Cor.   I.     (Converse   of  Prop.    X).     A  straight  line 
perpendicular  to  a  radius  at  its  outer  extremity  is  tangent 
to  the  circle. 

HINT.  Prove  by  the  indirect  method.  In  the  figure  for  Prop.  X, 
suppose  that  AB  is  not  tangent  to  circle  O  at  point  T]  then  draw  CD 
through  T,  tangent  to  circle  O.  Apply  §63. 

315.  Cor.  H.     A  perpendicular  to  a  tangent  at  the  point 
of  tangency  passes  through  the  center  of  the  circle. 


1. 

REASONS 
§  286. 

2. 

§  54,  15. 

3. 
4. 
5. 
6. 

§  54,  12. 
§  279,  a. 
§  309. 
Arg.  5. 

7. 

§165. 

BOOK  II  127 

316.  Cor.  HI.     A  line  drawn  from  the  center  of  a  circle 
perpendicular  to  a  tangent  passes  through  the  point  of 
tangency. 

317.  Def.     A  polygon  is  circum- 
scribed about  a  circle  if  each  side 
of  the  polygon  is  tangent  to  the 
circle.      In   the   same   figure   the 
circle  is  said  to  be  inscribed  in  the 
polygon. 

Ex.  437.  The  perpendiculars  to  the 
sides  of  a  circumscribed  polygon  at  their 
points  of  tangency  pass  through  a  com- 
mon point. 

Ex.  438.  The  line  drawn  from  any  vertex  of  a  circumscribed  polygon 
to  the  center  of  the  circle  bisects  the  angle  at  that  vertex  and  also  the 
angle  between  radii  drawn  to  the  adjacent  points  of  tangency. 

Ex.  439.  If  two  tangents  are  drawn  from  a  point  to  a  circle,  the 
bisector  of  the  angle  between  them  passes  through  the  center  of  the  circle. 

Ex.  440.  The  bisectors  of  the  angles  of  a  circumscribed  quadrilateral 
pass  through  a  common  point. 

Ex.  441.  Tangents  to  a  circle  at  the  extremities  of  a  diameter  are 
parallel. 

PROPOSITION  XI.     PROBLEM 

318.  To  construct  a  tangent  to  a  circle  at  any  given 
point  in  the  circumference. 

The  construction,  proof,  and  discussion  are  left  as  an  exer- 
cise for  the  student.  (See  §  314.) 


Ex.  442.  Construct  a  quadrilateral  which  shall  be  circumscribed  about 
a  circle.  What  kinds  of  quadrilaterals  are  circumscriptible  ? 

Ex.  443.  Construct  a  parallelogram  which  shall  be  inscribed  in  a 
circle.  What  kinds  of  parallelograms  are  inscriptible  ? 

Ex.  444.  Construct  a  line  which  shall  be  tangent  to  a  given  circle 
and  parallel  to  a  given  line. 

Ex.  445.  Construct  a  line  which  shall  be  tangent  to  a  given  circle 
and  perpendicular  to  a  given  line. 


128  PLANE   GEOMETRY 

319.  Def.    The  length  of  a  tangent  is  the  length  of  the  seg- 
ment included  between  the  point  of  tangency  and  the  point 
from  which  the  tangent  is  drawn;  as  TP  in  the  following  figure. 

PROPOSITION  XII.     THEOREM 

320.  If  two  tangents  are  drawn  from  any  given  point 
to  a  circle,  these  tangents  are  equal. 


Given   PT  and  PS,  two  tangents  from  point  P  to  circle  0. 

To  prove   PT=PS. 

The  proof  is  left  as  an  exercise  for  the  student. 


Ex.  446.  The  sum  of  two  opposite  sides  of  a  circumscribed  quadrilat- 
eral is  equal  to  the  sum  of  the  other  two  sides. 

Ex.  447.  The  median  of  a  circumscribed  trapezoid  is  one  fourth  the 
perimeter  of  the  trapezoid. 

Ex.  448.  A  parallelogram  circumscribed  about  a  circle  is  either  a 
rhombus  or  a  square. 

Ex.  449.  The  hypotenuse  of  a  right  triangle  circumscribed  about  a 
circle  is  equal  to  the  sum  of  the  other  two  sides  minus  a  diameter  of  the 
circle. 

Ex.  450.  If  a  circle  is  inscribed  in  any  triangle,  and  if  three  triangles 
are  cut  from  the  given,  triangle  by  drawing  tangents  to  the  circle,  then 
the  sum  of  the  perimeters  of  the  three  triangles  will  equal  the  perimeter 
of  the  given  triangle. 


BOOK   II 


129 


PROPOSITION  XIII.     PROBLEM 
321.     To  inscribe  a  circle  in  a  given  triangle. 

B 

H 


Given    A  ABC. 

To  inscribe  a  circle  in  A  AB  C. 


I.    Construction 

1.  Construct  AE  and  (7/>,  bisecting  A  CAB  and  BCA,  respec- 
tively.    §  127. 

2.  AE  and  CD  will  intersect  at  some  point  as  0.     §  194. 

3.  From  0  draw  OF  ±  AC.     §  149. 

4.  With  0  as  center  and  OF  as  radius  construct  circle  FGH. 

5.  Circle  FGH  is  inscribed  in  A  ABC. 

II.   The  proof  and  discussion  are  left  for  the  student. 

322.  Def.  A  circle  which  is  tangent  to 
one  side  of  a  triangle  and  to  the  other  two 
sides  prolonged  is  said  to  be  escribed  to  the 
triangle.  

Ex.  451.  Problem.  To  escribe  a  circle  to  a 
given  triangle. 

Ex.  452.     (a)  Prove  that  if  the  lines  that  bi- 
sect three  angles  of  a  quadrilateral  meet  at  a  com- 
mon point  P,  then  the  line  that  bisects  the  remaining  angle  of  the  quadri- 
lateral passes  through  P.    (&)    Tell  why  a  circle  can  be  inscribed  in  this 
particular  quadrilateral. 

Ex.  453.     In  triangle  ABC,  draw  XY  parallel  to  BC  so  that 
XY  +  BC=BX+  CT. 

Ex.  454.     Inscribe  a  circle  in  a  given  rhombus. 


130  PLANE   GEOMETRY 

PROPOSITION   XIV.     PROBLEM 
323.   To  circumscribe  a  circle  about  a  given  triangle. 


\ 

Given    AAHC. 

To  circumscribe   a  circle  about  A  ABC. 

The  construction,  proof,  and  discussion  are  left  as  an  exercise 
for  the  student. 

324.  Cor.  Three  points  not  in  tlw  same  straight  line 
determine  a  circle.  

Ex.  455.  Discuss  the  position  of  the  center  of  a  circle  circumscribed 
about  an  acute  triangle  ;  a  right  triangle  ;  an  obtuse  triangle. 

Ex.  456.     Circumscribe  a  circle  about  an  isosceles  trapezoid. 

Ex.  457.  Given  the  base  of  an  isosceles  triangle  and  the  radius  of 
the  circumscribed  circle,  to  construct  the  triangle. 

Ex.  458.  The  inscribed  and  circumscribed  circles  of  an  equilateral 
triangle  are  concentric. 

Ex.  459.  If  upon  the  sides  of  any  triangle  equilateral  triangles  are 
drawn,  and  circles  circumscribed  about  the  three  triangles,  these  circles 
will  intersect  at  a  common  point. 

Ex.  460.  The  two  segments  of  a  secant  which  are  between  two  con- 
centric circumferences  are  equal. 

Ex.  461.  The  perpendicular  bisectors  of  the  sides  of  an  inscribed 
quadrilateral  pass  through  a  common  point. 

Ex.  462.  The  bisector  of  an  arc  of  a  circle  is  determined  by  the  center 
of  the  circle  and  another  point  equidistant  from  the  extremities  of  the 
chord  of  the  arc. 

Ex.  463.  If  two  chords  of  a  circle  are  equal,  the  lines  which  connect 
their  mid-points  with  the  center  of  the  circle  are  equal. 


BOOK  II 


131 


TWO   CIRCLES 

325.  Def.     The    line   determined    by    the   centers   of  two 
circles  is  called  their  line  of  centers  or  center-line. 

326.  Def.     Concentric    circles   are  circles   which   have   the 
same  center. 

PROPOSITION   XV.     THEOREM 

327.  If  two  circumferences  meet  at  a  point  which  is  not 
on  their  line  of  centers,  they  also  meet  in  one  other  point. 

M 

N 


Given   circumferences  M  and  N  meeting  at  P,  a  point  not  on 
their  line  of  centers  OQ. 

To  prove   that  the  circumferences  meet  at  one  other  point, 

as  R. 

ARGUMENT  REASONS 

1.  Draw  OP  and  QP.  1.    §  54,  15. 

2.  Rotate  A  OPQ  about  OQ  as  an  axis  until      2.    §  54,  14. 

it  falls  in  the  position  QRO. 

3.  OR  =  OP  —  a  radins  of  circle  M. 

4.  .-.  R  is  on  circumference  M. 


5.  Also  QR  =  QP  =  a  radius  of  circle  N. 

6.  .*.  R  is  on  circumference  N. 

7.  .*.  R  is  on  both  circumference   M  and 

circumference  JV;  i.e.  circumferences 
M  and  N  meet  at  R.  Q.E.D. 


3.  By  cons. 

4.  §  279,  a. 

5.  By  cons. 

6.  §  279,  a. 

7.  Args.  4  and  6. 


328.   Cor.  I.     If  two  circumferences  intersect,  their  line 
of  centers  bisects  their  common  chord  at  right  angles. 


132 


PLANE   GEOMETRY 


329.  Cor.  II.     If  two  circumferences  meet  at  one  point 
only,  tJicut  point  is  on  their  line  of  centers. 

HINT.     If  they  meet  at  a  point  which  is  not  on  their  line  of  centers, 
they  also  meet  in  another  point  (§  327).     This  contradicts  the  hypothesis. 

330.  Def .    Two  circles  are  said  to  touch  or  be  tangent  to  each 
other  if  they  have  one  and  only  one  point  in  common.     They 
are  tangent  internally  or  externally  according  as  one  circle  lies 
within  or  outside  of  the  other. 


Tangent  externally.  Tangent  internally.      .          Concentric. 

331.  From  §  330,  Cor.  II  may  be  stated  as  follows : 

If  two  circles  are  tangent  to  each  other,  their  common  point  lies 
on  their  line  of  centers. 

332.  Cor.  III.  If  two  circles  are  tangent  to  each  other, 
they  have  a  common  tangent  line  at  their  point  of  contact. 

HINT.     Apply  §  314. 

333.  Def.     A  line  touching  two  circles  is  called  an  external 
common  tangent  if  both  circles  lie  on  the  same  side  of  it ;  the 
line  is  called  an  internal  common  tangent  if  the  two  circles  lie 
on  opposite  sides  of  it. 


FIG.  1. 


FIG.  2. 


Thus  a  belt  connecting  two  wheels  as  in  Fig.  1  is  an  illus- 
tration of  external  common  tangents,  while  a  belt  arranged  as 
in  Fig.  2  illustrates  internal  common  tangents. 


BOOK  II  133 

334.  Questions.  In  case  two  circles  are  tangent  internally  how 
many  common  tangents  can  be  drawn  ?  in  case  their  circumferences 
intersect  ?  in  case  they  are  tangent  externally  ?  in  case  they  are 
wholly  outside  of  each  other  ?  in  case  one  is  wholly  within  the  other  ? 


Ex.  464.  If  two  circles  intersect,  their  line  of  centers  bisects  the 
angles  between  the  radii  drawn  to  the  points  of  intersection. 

Ex.  465.  If  the  radii  of  two  intersecting  circles  are  5  inches  and 
8  inches,  what  may  be  the  length  of  the  line  joining  their  centers? 

Ex.  466.  If  two  circles  are  tangent  externally,  tangents  drawn  to 
them  from  any  point  in  their  common  internal  tangent  are  equal. 

Ex.  467.  Two  circles  are  tangent  to  each  other.  Construct  their  com- 
mon tangent  at  their  point  of  contact. 

Ex.  468.  Construct  a  circle  passing  through  a  given  point  and  tan- 
gent to  a  given  circle  at  another  given  point. 

Ex.  469.  Find  the  locus  of  the  centers  of  all  circles  tangent  to  a  given 
circle  at  a  given  point. 

MEASUREMENT 

335.  Def.     To  measure  a  quantity  is  to  find  how  many  times 
it  contains  another  quantity  of  the  same  kind.     The  result  of 
the    measurement  is   a  number    and    is    called    the    numerical 
measure,  or  measure-number,  of  the  quantity  which  is  measured. 
The  measure  employed  is  called  the  unit  of  measure. 

Thus,  the  length  or  breadth  of  a  room  is  measured  by  find- 
ing how  many  feet  there  are  in  it ;  i.e.  how  many  times  it  con- 
tains a  foot  as  a  measure. 

336.  It  can  be  shown  that  to  every  geometric   magnitude 
there  corresponds  a  definite  number  called  its  measure-number. 
The  proof  that  to  every  straight  line  segment  there  belongs  a 
measure-number  is  found  in  the  Appendix,  §  595.     The  method 
of  proof  there  used  shows  that  operations  with  measure-numbers 
follow  the  ordinary  laws  of  algebra. 

337.  Def.     Two    quantities    are     commensurable    if     there 
exists  a  measure   that   is   contained   an   integral   number   of 
times  in  each.     Such  a  measure  is  called  a  common  measure 
of  the  two  quantities, 


134  PLANE   GEOMETRY 

Thus,  a  yard  and  a  foot  are  commensurable,  each  containing 
an  inch  a  whole  number  of  times;  so.  too,  12|  inches  and 
18}  inches  are  commensurable,  each  containing  a  fourth  of  an 
inch  a  whole  number  of  times. 

338.  Questions.     If  two   quantities  have  a  common  measure,  hoio 
many  common  measures  have  they  ?    Name  some  common  measures  of 
12^  inches  and  18|  inches.     What  is  their  greatest  common  measure  ? 
What  is  their  least  common  measure  ?  \ 

339.  Def.     Two   quantities   are  incommensurable   if    there 
exists  no  .measure  that  is  contained  an  integral  number  of 
times  in  each. 

It  will  be  shown  later  that  a  diagonal  and  a  side  of  the  same 
square  cannot  be  measured  by  the  same  unit,  without  a  re- 
mainder; and  that  the  diagonal  is  equal  to  V2  times  the 
numerical  measure  of  the  side.  Now  V2  can  be  expressed 
only  approximately  as  a  simple  fraction  or  as  a  decimal.  It 
lies  between  1.4  and  1.5,  for  (1.4)2  =  1.96,  and  (1.5)2  =  2.25. 
Again,  it  lies  between  1.41  and  1.42,*  between  1.414  and  1.415, 
between  1.4142  and  1.4143,  and  so  on.  By  repeated  trials 
values  may  be  found  approximating  more  and  more  closely 
to  V2,  but  no  decimal  number  can  be  obtained  that,  taken  twice 
as  a  factor,  will  give  exactly  2. 

340.  When   we   speak   of    the    ratio   of    one    quantity   to 
another,  we  have  in  mind  their  relative  sizes.     By  this  is  meant 
not  the  difference  between  the  two,  but  how  many  times  one  con- 
tains the  other  or  some  aliquot  part  of  it.     In  algebra  the  ratio 
of  two  numbers  has  been  defined  as  the  indicated  quotient  of 
the  first  divided  by  the  second.     Since  to  each  geometric  mag- 
nitude there  corresponds  a  number  called  its  measure-number 
(§  336),  therefore : 

341.  Def.     The  ratio  of  two  geometric  magnitudes  may  be 
defined  as  the  quotient  of  their  measure-numbers,  when  the 
same  measure  is  applied  to  each. 

*  The  student  should  multiply  to  get  the  successive  approximations. 


BOOK  II  135 

Thus,  if  the  length  of  a  room  is  36  feet  and  the  width  27  feet, 
the  ratio  of  the  length  to  the  width  is  said  to  be  the  ratio  of 
36  to  27 ;  i.e.  f  f ,  which  is  equal  to  f .  The  ratio  of  the  width 
to  the  length  is  f  J,  which  is  equal  to  J.  The  term  ratio  is 
never  applied  to  two  magnitudes  that  are  unlike. 

342.  Def.     If  the  two  magnitudes  compared  are  commen- 
surable, the  ratio  is   called  a  commensurable  ratio   and   can 
always  be  expressed  as  a  simple  fraction. 

343.  Def.     If  the  two  magnitudes  compared  are  incommen- 
surable, the  ratio  is  called  an  incommensurable  ratio  and  can 
be  expressed  only  approximately  as  a  simple  fraction.     Closer 
and  closer  approximations  to  an  incommensurable  ratio  may 
be  obtained  by  repeatedly  using  smaller  and  smaller  units  as 
measures  of  the  two  magnitudes  to  be  compared  and  by  finding 
the  quotient  of  the  numbers  thus  obtained. 

Two  magnitudes,  e.g.  two  line  segments,  taken  at  random  are 
usually  incommensurable,  commensurability  being  compara- 
tively rare. 

344.  Historical  Note.     The  discovery  of  incommensurable  magni- 
tudes is  ascribed  to  Pythagoras,  whose  followers  for  a  long  time  kept  the 
discovery  a  secret.     It  is  believed  that  Pythagoras  was  the  first  to  prove 
that  the  side  and  diagonal  of  a  square  are  incommensurable.    A  more 
complete  account  of  the  work  of  Pythagoras  will  be  found  in  §  510. 


Ex.  470.  What  is  the  greatest  common  measure  of  48  inches  and 
18  inches  ?  Will  it  divide  48  inches  —  18  inches  ?  48  inches  —  2  x  18 
inches  ? 

Ex.  471.  Draw  any  two  line  segments  which  have  a  common  meas- 
ure. Find  the  sum  of  these  lines  and,  by  laying  off  the  common  measure, 
show  that  it  is  a  measure  of  the  sum  of  the  lines. 

Ex.  472.  Given  two  lines,  5  inches  and  4  inches  long,  respectively. 
Show  by  a  diagram  that  any  common  measure  of  5  inches  and  4  inches  is 
also  a  measure  of  15  inches  plus  or  minus  8  inches. 

Ex.  473.  Find  the  greatest  common  divisor  of  728  and  844  by  division 
and  point  out  the  similarity  of  the  process  to  that  used  in  Prop.  XVI. 


136  PLANE   GEOMETRY 

PROPOSITION  XVI.     PROBLEM 

345.  To  determine  whether  two  given  lines  are  commen- 
surable or  not;  and  if  they  are  commensurable,  to  find 
their  common  measure  and  their  ratio. 


F  B 


E  G  D 


Given   lines  AB  and  CD. 

To  determine:     (a)  whether  AB  and  CD  are  commensurable 

and  if  so, 

(6)  what  is  their  common  measure;  and 
(c)  what  is  the  ratio  of  AB  to  CD. 

I.    Construction 

1.  Measure  off  AB  on  CD  as  many  times  as  possible.     Sup- 
pose it  is  contained  once,  with  a  remainder  ED. 

2.  Measure  off  ED  on  AB  as  many  times  as  possible.     Sup- 
pose it  is  contained  twice,  with  a  remainder  FB. 

3.  Measure  off  FB  on  ED  as  many  times  as  possible.     Sup- 
pose it  is  contained  three  times,  Avith  a  remainder  GD. 

4.  Measure  off  GD  on  FB  as  many  times  as  possible,  and  so  on. 

5.  It  is  evident  that  this  process  will  terminate  only  when 
a  remainder  is  obtained  which  is  a  measure  of  the  remainder 
immediately  preceding. 

6.  If  this  process  terminates,  then  the  two  given  lines  are 
commensurable,  and  the  last  remainder  is  their  greatest  com- 
mon measure. 

7.  For  example,  if  GD  is  a  measure  of  FB,  then  AB  and  CD 
are  commensurable,  GD  is  their  greatest  common  measure,  and 
the  ratio  of  AB  to  CD  can  be  found. 


II.    Proof 


ARGUMENT 


1.  Suppose  FB  =  2  GD. 

2.  ED  =  EG  +  GD. 


REASONS 

1.  See  I,  7. 

2.  §54,11. 


BOOK   II 


137 


3.  /.  ED  =  3  FB  +  GD  =  1  GD. 

4.  AB  —  2  ED "+  FB. 

5.  .'.  AB  =  14  GD  +  2  GD  =  16  S.D. 

6.  CD  ==  .45  -f  £7). 

7.  .•.  C/>  =16  G/>  4-  7  (?Z)  =  23  GZ>. 

8.  .-.  ^45  and  CD  are  commensurable. 

9.  Also,  GD  is  a  common  measure  of  AB       9.    Args.  5  and 

and  CD. 

LO.    The  ratio  of  AB  to  CD  =  the  ratio  of 
16  GD  to  23  GD  =  if.  Q.E.D. 

III.    A  full  discussion  of  this  problem  will  be  found  in  the 
Appendix,  §  598. 


3. 

§309. 

4. 

§  54,  11. 

5. 

§309. 

6. 

§  54,  11. 

7. 

§309. 

8. 

§337. 

9. 

Args.  5 

7. 

10. 

§  341. 

CONSTANTS   AND   VAKIABLES.     LIMITS 

346.  Consider  an  isosceles  triangle  ABC,  whose  base  is  AC 
and  whose  altitude  is  LB.  Keeping  the  base  AC  the  same  (con- 
stant), suppose  the  altitude  to  change  (vary). 

If  LB   increases,   what   will    be   the  B 

effect  upon  the  lengths  of  AB  and  CB  ? 
what  the  effect  upon  the  base  angles  ? 
upon  the  vertex  angle  ?  Will  the  base 
angles  always  be  equal  to  each  other  ? 
What  limiting  value  have  they  ?  Is  the 
base  angle  related  to  half  the  vertex 
angle  or  are  the  two  independent  ? 

What  relation  is  there  ?    Is  this  relation 

A  L  C 

constant  or  does  it  change  ? 

Imagine  the  altitude  of  the  triangle  to  diminish.  Repeat  the 
questions  given  above,  considering  the  altitude  as  decreasing. 
What  is  now  the  limiting  value  for  the  altitude  ?  what  for  the 
length  of  one  of  the  equal  sides  ?  for  the  base  angles  ?  for  the 
angle  at  the  vertex  ? 

Ex.  474.  Consider  an  isosceles  triangle  with  a  constant  altitude  and 
a  variable  base.  Repeat  the  questions  given  above. 


138  PLANE   GEOMETRY 

Ex.  475.  Consider  an  isosceles  triangle  with  constant  base  angles, 
but  variable  base.  Tell  what  other  constants  and  what  other  variables 
there  would  be  in  this  case. 

Ex.  476.  If  through  any  point  in  the  base  of  an  isosceles  triangle 
lines  are  drawn  parallel  to  the  equal  sides  of  the  triangle,  a  parallelo- 
gram will  be  formed  whose  perimeter  will  be  constant ;  i.e.  the  perimeter 
will  be  independent  of  the  position  of  the  point. 


347.  Def.     A  magnitude  is  constant  if  it  does  not  change 
throughout  a  discussion. 

348.  Def.     A  magnitude  is  variable  if  it  takes  a  series  of 
different  successive  values  during  a  discussion. 

349.  Def.     If  a  variable  approaches  a  constant  in  such  a  way 
that  the  difference  between  the  variable  and  the  constant  may 
be  made  to  become  and  remain  smaller  than  any  fixed  number 
previously  assigned,  however  small,  the  constant  is  called  the 
limit  of  the  variable. 

350.  The  variable  is  said  to  approach  its  limit  as  it  becomes 
more  and  more  nearly  equal  to  it.     Thus,  suppose  a  point  to 
move  from  A  toward  B,  by 

successive  steps,  under  the       A £ ¥     ?*    ? 

restriction    that     at     each 

step  it  must  go  over  one  half  the  segment  between  it  and  B.  At 
the  first  step  it  reaches  C,  whereupon  there  remains  the  segment 
CB  to  be  traveled  over  ;  at  the  next  step  it  reaches  D,  and  there 
remains  an  equal  segment  to  be  covered.  Whatever  the  number 
of  steps  taken,  there  must  always  remain  a  segment  equal  to 
the  segment  last  covered.  But  the  segment  between  A  and  the 
moving  point  may  be  made  to  differ  from  AB  by  as  little  as  we 
please,  i.e.  by  less  than  any  previously  assigned  value.  For  assign 
some  value,  say,  half  an  inch.  Then  the  point,  continuing  to 
move  under  its  governing  law,  may  approach  B  until  there 
remains  a  segment  less  than  half  an  inch.  Whatever  be  the 
value  assigned,  the  variable  segment  from  A  to  the  moving 
point  may  be  made  to  differ  from  the  constant  segment  AB  by 
less  than  the  assigned  value. 


BOOK  II  139 

Again,  the  numbers  in  the  series  4,  2,  1,  ^,  J,  i,  etc.,  in  which 
each  term  is  one  half  of  the  preceding  term,  approach  0  as  a 
limit  as  the  number  of  terms  in  the  series  is  increased.  For  if 
we  assign  any  value,  as  Twrnr*  ^  *s  evident  that  a  term  of  the 
series  may  be  found  which  is  less  than  10j,00 ;  it  is  also  evident 
that  no  term  of  the  series  can  become  0. 

351.  In  elementary  geometry  the  variables  that   approach 
limits  are  usually  such  that  they  cannot  attain  their  limits. 
There  are,  however,  variables  that  do  attain  their  limits.     The 
limiting  values  of  algebraic  expressions  are  frequently  of  this 

kind ;  thus,  the  expression  — — -  approaches  1  as  x  approaches 

ar  -f- 1 

0,  and  has  the  limit  1  when  x  becomes  zero. 

352.  Def.     Two  variables  are  said  to  be  related  when  one 
depends  upon  the  other  so  that,  if  the  value  of  one  variable  is 
known,  the  value  of  the  other  can  be  obtained. 

For  example,  the  diagonal  and  the  area  of  a  square  are  re- 
lated variables,  for  there  is  a  value  for  the  area  for  any  value 
which  may  be  given  to  the  diagonal,  and  vice  versa. 

353.  Questions.     On  the  floor  is  a  bushel  of  sand.    If  we  keep  adding 
to  this  pile  forever,  how  large  will  it  become  ?     Does  it  depend  upon  the 
law  governing  our  additions  ?     If  we  add  one  quart  each  hour,  how  large 
will  it  become  ?     If  we  add  one  quart  the  first  hour,  a  half  quart  the  second 
hour,  a  fourth  quart  the  third  hour,  etc.,  each  hour  adding  one  half  as 
much  as  the  preceding  hour,  how  large  will  the  pile  become  ? 

354.  Historical  Note.     Achilles  and  the  Tortoise.     One  of  the 

early  Greek  schools  of  mathematics,  founded  during  the  fifth  century 
B.C.,  at  Elea,  Italy,  and  known  as  the  Eleatic  School,  was  famous  for 
its  investigations  of  problems  involving  infinite  series.  Zeno,  one  of  the 
most  prominent  members,  proposed  this  question:  He  "argued  that  if 
Achilles  ran  ten  times  as  fast  as  a  tortoise,  yet  if  the  tortoise  had  (say) 
1000  yards  start,  it  could  never  be  overtaken  :  for,  when  Achilles  had 
gone  the  1000  yards,  the  tortoise  would  still  be  100  yards  in  front  of  him ; 
by  the  time  he  had  covered  these  100  yards,  it  would  still  be  10  yards  in 
front  of  him  ;  and  so  on  forever :  thus  Achilles  would  get  nearer  and 
nearer  to  the  tortoise  but  never  overtake  it."  Was  Zeno  right  ?  If  not, 
can  you  find  the  fallacy  in  his  argument  ? 


140 


PLANE   GEOMETRY 


PROPOSITION  XVII.     THEOREM 

355.   If  two  variables  are  always  equal,  and  if  each  ap- 
proaches a  limit,  then  their  limits  are  equal. 

Given   two  variables,  V  and  F',  which  are  always  equal  and 
which  approach  as  limits  L  and  L',  respectively. 
To  prove    L  =  L1. 


ARGUMENT 

1.  Either  L  =  L',  or  L  =£  L1. 

2.  Suppose  that  one  limit  is  greater  than 

the  other,  say  L  >  L' ;  then  F,  in  ap- 
proaching L,  may  assume  a  value  be- 
tween L'  and  L\  i.e.  V  may  assume  a 
value  >  L'. 

3.  But  F'  cannot  assume  a  value  >  L'. 

4.  .-.   F  may  become  >  F'. 

5.  But  this  is  impossible,  since  F  and  F' 

are  always  equal. 

L  =  L'.  Q.E.D. 


6. 


REASONS 

1.  §161,  a. 

2.  §  349. 


§  349. 

Args.  2  and  3. 


5.  By  hyp. 

6.  §  161,  6. 


356.  Question.  In  the  above  proof  are  V  and  V  increasing  or  de- 
creasing variables  ?  The  student  may  adapt  the  argument  above  to  the 
case  in  which  V  and  V  are  decreasing  variables. 


Ex.  477.  Apply  Prop.  XVII 
to  the  accompanying  figure, 
where  variable  V  is  represented 
by  the  line  AB,  variable  V  by 
the  line  CD,  limit  L  by  line  AE, 
and  limit  L'  by  line  CF. 


B 


H 


357.    Note.     It  will  be  seen  that,  in  the  application  of  Prop.  XVII, 
there  are  three  distinct  things  to  be  considered : 

(1)  Two  variables  that' are  always  equal; 

(2)  The  limits  of  these  two  variables ; 

(3)  The  equality  of  these  limits  themselves,. 


BOOK  II 


141 


PROPOSITION  XVIII.     THEOREM 

358.   An  angle  at  the  center  of  a  circle  is  measured 
its  intercepted  arc. 


Given  central  /-AOB  and  AB  intercepted  by  it;  let  /.COE  be 
any  unit  Z  (e.g.  a  degree),  and  let  CE,  intercepted  by  the  unit  Z, 
be  the  unit  arc. 

To  prove  the  measure-number  of  /.AOB,  referred  to  /.COE, 
equal  to  the  measure-number  of  AB,  referred  to  CE. 

I.   If  /.AOB  and  /.COE  are  commensurable. 

(a)  Suppose  that  /.COE  is  contained  in  /.AOB  an  integral 
number  of  times. 


ARGUMENT 

1.  Apply  Z  COE  to  /.AOB  as   a  measure. 

Suppose  that  /iCOE  is  contained  in 
Z  AOB  r  times. 

2.  Then  r  is  the  measure-number  of  Z  ^10,8 

referred  to  /.COE  as  a  unit. 

3.  Now  the  r  equal  central  A  which  com- 

pose /.AOB  intercept  r  equal  arcs  on 
the  circumference,  each  equal  to  CE. 

4.  .-.  r  is  the  measure-number  of  AB  re- 

ferred to  CE  as  a  unit. 

5.  .-.  the    measure-number    of   /.AOB,   re- 

ferred to  /.COE  as  a  unit,  equals  the 
measure-number  of  AB,  referred  to 
CE  as  a  unit.  Q.E.D. 


REASONS 

1.  §  335. 

2.  §  335. 

3.  §  293,  I. 

4.  §335. 

5.  §54,1. 


142 


PLANE   GEOMETRY 


I.  It  Z.AOB  and  Z  COE  are  commensurable. 

(6)  Suppose  that  Z  COE  is  not  contained  in  /.AOB  an  integral 
number  of  times.  The  proof  is  left  as  an  exercise  for  the 
student. 

HINT.  Some  aliquot  part  of  ZCOE  must  be  a  measure  of  /.AOB. 
(Why  ?)  Try  \  <S_COE,  $  ZCOE,  etc. 

II.  If  Z  AOB  and  Z  COE  are  incommensurable. 


1. 


3. 


ARGUMENT 

Let  Z  1  be  a  measure  of  Z  COE.  Ap- 
ply Z  1  to  Z.  AOB  as  many  times  as 
possible.  There  will  then  be  a  re- 
mainder, /-FOB,  less  than  Zl. 

Z  AOF  and  Z  (70.E  are  commensurable. 

.-.  the  measure-number  of  /.AOF,  re- 
ferred to  Z  COE  as  a  unit,  equals  the 
measure-number  of  AF,  referred  to 
CE  as  a  unit. 


REASONS 
1.    §339. 


§337. 
§  358,  I. 


BOOK   II 


143 


ARGUMENT  REASONS 

4.  Now  take  a  smaller  measure  of  /.COE.        4.    §  335. 

No  matter  how  small  a  measure  of 
Z  COE  is  taken,  when  it  is  applied  as 
a  measure  to  Z  AOB,  the  remainder, 
Z  FOB,  will  be  smaller  than  the  Z 
taken  as  a  measure. 

5.  Also  FB  will  be  smaller  than  the  arc  in-        5.    §  294. 

tercepted  by  the  Z  taken  as  a  measure. 

6.  .-.  the  difference  between  /.AOF  and        6.   Args.4and5. 

Z  AOB  may  be  made  to  become  and 
remain  less  than  any  previously  as- 
signed Z,  however  small ;  and  like- 
wise the  difference  between  AF  and 
AB,  less  than  the  arc  intercepted  by 
the  assigned  Z. 

7.  .-.  Z  AOF  approaches  Z  AOB  as  a  limit,       7.    §  349. 

and  AF  approaches  AB  as  a  limit. 

8.  Hence  the  measure-number  of  Z  AOF       8.    §  359. 

approaches  the  measure-number  of 
/.AOB  as  a  limit,  and  the  measure- 
number  of  AF  approaches  the  meas- 
ure-number of  AB  as  a  limit. 

9.  But  the   measure-number  of  /.AOF  is        9.    Arg.  3. 

always  equal  to  the  measure-number 
of  AF. 

10.    .-.  the   measure-number  of  /.AOB,  re-      10-  .§355. 
ferred  to  Z  COE  as  a  unit,  equals  the 
measure-number  of  AB,  referred  to 
CE  as  a  unit.  Q.E.D. 

359.  If  a  magnitude  is  variable  and  approaches  a  limit,  then, 
as  the  magnitude  varies,  the  successive  measure-numbers  of  the 
variable  approach  as  their  limit  the  measure-number  of  the  limit 
of  the  magnitude. 

(This  theorem  will  be  found  in  the  Appendix,  §  597.) 


144  PLANE   GEOMETRY 

360.  Cor.  In  equal  circles,  or  in  the  same  circle,  two 
angles  at  the  center  have  the  same  ratio  as  their  inter- 
cepted arcs. 

HINT.  The  measure-numbers  of  the  angles  are  equal  respectively  to 
the  measure- numbers  of  their  intercepted  arcs.  Therefore  the  ratio  of 
the  angles  is  equal  to  the  ratio  of  the  arcs. 


Ex.  478.  Construct  a  secant  which  shall  cut  off  two  thirds  of  a  given 
circumference. 

Ex.  479.  Is  the  ratio  of  two  chords  in  the  same  circle  equal  to  the 
ratio  of  the  arcs  which  they  subtend  ?  Illustrate  your  answer,  using  a 
semicircumference  and  a  quadrant. 


361.  JThe  symbol  oc  will  be  used  for  is  measured  b^     cc  is 
the  symbol  of  variation,  and  the  macron  (— )  means  long  or 
length.     Hence  QC  suggests  varies  as  the  length  of. 

362.  From  §  336  it  follows  directly  that : 

(a)  In  equal  circles,  or  in  the    same  circle,  equal  angles  are 
measured  *  by  equal  arcs  ;   conversely,  equal  arcs  measure  equal 
a,ngles.  ( 

\  Sum, 

(b)  The  measure  of  the  \  [of  two  angles  is  equal  to 

the  \   ,  \  of  the  measures  of  the  angles. 

{  difference  j 

(c)  TJie  measure  of  any  'multiple,  of  an  angle  is  equal  to  that 
same  multiple  of  the  measure  of  the  angle. 

363.  Def.     An  angle  is  said  to  be  inscribed  in  a  circle  if  its 
vertex  lies  on  the  circumference  and  its  sides  are  chords. 

354.  Def.  An  angle  is  said  to  be  inscribed  in  a  segment  of 
a  circle  if  its  vertex  lies  on  the  arc  of  the  segment  and  its 
sides  pass  through  the  extremities  of  that  arc. 

*Itis,  of  course,  inaccurate  to  speak  of  measuring  one  magnitude  by  another  magnitude 
of  a  different  kind ;  but,  in  this  case,  it  has  become  a  convention  so  general  that  the 
student  needs  to  become  familiar  with  it.  More  accurately,  in  Prop.  XVIII,  the  measure- 
number  of  an  angle  at  the  center,  referred  to  any  unit  angle,  is  the  same  as  the  meaaure- 
number  of  its  intercepted  arc  when  the  unit  arc  is  the  arc  intercepted  by  the  unit  angle. 


BOOK  II 


145 


PROPOSITION   XIX.     THEOREM 

365.   An  inscribed  angle  is  measured  by  one  half  its 
intercepted  arc. 

B  B  B 


Given   inscribed  Z  ABC. 

To  prove    that  Z  ABC  oc  1  AC. 

I.    Let  one  side  of  Z  ABC,  as  AB,  pass  through  the  center  of 
the  circle  (Fig.  1). 

REASONS 

1.  §  54,  15. 

2.  §279,  a. 

3.  §111. 

4.  §215. 

5.  §309. 

6.  §  54,  8  a. 

7.  §358. 


ARGUMENT 

1.  Draw  radius  OC. 

2.  OB  =  OC. 

3.  .-.  Z1  =  Z2. 

4.  Z1  +  Z2  =  Z3. 

5.  .-.  2Z1  =  Z3. 

6.  .-.  Zl  =  iZj3. 

7.  But  Z  3  oc  U. 

8.  .-.  J-  Z  3  or  Z  1  *  1.  AC- 

L  e.  Z  ^£(7  <X\AC. 

II.    Let  center  0  lie  within  Z  ^LBtf  (Fig.  2). 
III.   Let  center  0  lie  outside  of  Z  ^LBGY  (Fig.  3). 
The  proofs  of  II  and  III  are  left  as  exercises  for  the  student. 

HINT.  In  Fig.  2,  what  is  the  measure  of  Z4  ?  of  Z  5  ?  What,  then, 
is  the  measure  of  Z  A  BC?  In  Fig.  3,  what  is  the  measure  of  Z.XBCt 
of  Z  XBA  ?  What,  then,  is  the  measure  of  Z  ABC  ? 


Q.E.D. 


8.    §  362,  c. 


146 


PLANE   GEOMETRY 
#> 


FIG.  1. 


FIG.  2. 


FIG.  3. 


366.  Cor.  I.    All  angles  inscribed  in  the  same  segment 
are  equal.    (See  Fig.  1.) 

367.  Cor.  II.    Any  angle  inscribed  in  a  semicircle  is  a 
right  angle.    (See  Fig.  2.) 

368.  Cor.   III.    The  locus  of  the  vertex  of  a  right  tri- 
angle having  a  given  hypotenuse  as  base  is  the  circuin- 
ference  having  the  same  hypotenuse  as  diameter. 

369.  Cor.  IV.    Any  angle  inscribed  in  a  segment  less 
than  a  semicircle  -is  an  obtuse  angle.    (See  Fig.  3.) 

370.  Cor.  V.    Any  angle  inscribed  in  a  segment  greater 
than  a  semicircle  is  an  acute  angle.    (See  Fig.  3.) 


Ex.  480.  If  an  inscribed  angle  contains  24  angle  degrees,  how  many 
arc  degrees  are  there  in  the  intercepted  arc  ?  how  many  in  the  rest  of 
the  circumference  ? 

Ex.  481.  If  an  inscribed  angle  intercepts  an  arc  of  70°,  how  many 
degrees  are  there  in  the  angle  ? 

Ex.  482.  How  many  degrees  are  there  in  an  angle  inscribed  in  a  seg- 
ment whose  arc  is  140°  ? 

Ex.  483.  Construct  any  segment  of  a  circle  so  that  an  angle  inscribed 
in  it  shall  be  an  angle  ^of  :  (a)  60°  ;  (ft)  45° ;  (c)  30°. 

Ex.  484.  Repeat  Ex.  483,  using  a  given  line  as  chord  of  the  segment. 
How  many  solutions  are  there  to  each  case  of  Ex.  483  ?  how  many  to 
each  case  of  Ex.  484  ? 

Ex.  485.  The  opposite  angles  of  an  inscribed  quadrilateral  are  sup- 
plementary. 


BOOK  II 


147 


Ex.  486.     If  the  diameter  of  a  circle  is  one  of  the  equal  sides  of  an 
isosceles  triangle,  the  circumference  will  bisect  the  base  of  the  triangle. 

Ex.  487.     By  means  of  a  circle  construct  a  right  triangle,  given  the 
hypotenuse  and  an  arm. 

Ex.  488.     By  means  of  a  circle  construct  a  right  triangle,  given  the 
hypotenuse  and  an  adjacent  angle. 

Ex.  489.     Construct  a  right  triangle,  having  given  the  hypotenuse  and 
the  altitude  upon  the  hypotenuse. 


371.     Historical  Note.     Thales  (640-546  B.C.),  the  founder  of  the 

earliest  Greek  school  of  niathe-  

matics,  is  said  to  have  discov- 
ered that  all  triangles  having 
a  diameter  of  a  circle  as  base, 
with  their  vertices  on  the  cir- 
cumference, have  their  vertex 
angles  right  angles.  Thales 
was  one  of  the  Seven  Wise 
Men.  He  had  much  business 
shrewdness  and  sagacity,  and 
was  renowned  for  his  practi- 
cal and  political  ability.  He 
went  to  Egypt  in  his  youth, 
and  while  there  studied  geom- 
etry and  astronomy.  The 
story  is  told  that  one  day  while 
viewing  the  stars,  he  fell  into 

a  ditch  ;  whereupon  an  old  woman  said,  "  How  canst  thou  know  what  is 
doing  in  the  heavens,  when  thou  seest  not  what  is  at  thy  feet  ?  " 

According  to  Plutarch,  Thales  computed  the  height  of  the  Pyramids  of 
Egypt  from  measurements  of  their  shadows.  Plutarch  gives  a  dialogue  in 
which  Thales  is  addressed  thus,  "Placing  your  stick  at  the  end  of  the 
shadow  of  the  pyramid,  you  made  by  the  same  rays  two  triangles,  and 
so  proved  that  the  height  of  the  pyramid  was  to  the  length  of  the  stick 
as  the  shadow  of  the  pyramid  to  the  shadow  of  the  stick."  This  compu- 
tation was  regarded  by  the  Egyptians  as  quite  remarkable,  since  they  were 
not  familiar  with  applications  of  abstract  science. 

The  geometry  of  the  Greeks  was  in  general  ideal  and  speculative,  the 
Greek  mind  being  more  attracted  by  beauty  and  by  abstract  relations 
than  by  the  practical  affairs  of  everyday  life. 


THALES 


148 


PLANE   GEOMETRY 


PROPOSITION  XX.     PROBLEM 

372.    To  construct  a  perpendicular  to  a  given  straight 
line  at  a  given  point  in  the  line. 

(Second  method.     For  another  method,  see  §  148.) 


\ 


M 


/      ^ 

++' 

1 

\      ^ 

1 
1 

\  ^"" 

1 

\ 


Given  line  AB  and  P,  a  point  in  AB. 
To  construct   a  _L  to  AB  at  P. 

I.    Construction 

1.  With  O,  any  convenient  point  outside  of  AB,  as  center, 
and  with  OP  as  radius,  construct  a  circumference  cutting  AB 
at  P  and  Q. 

2.  Draw  diameter  QM. 

3.  Draw  PM. 

4.  PM  is  J_  AB  at  P. 

II.    The  proof  and  discussion  are  left  as  an  exercise  for  the 
student. 

The  method  of  §  372  is  useful  when  the  point  P  is  at  or 
near  the  end  of  the  line  AB. 


Ex.  490.  Construct  a  perpendicular  to  line  AB  at  its  extremity  B, 
without  prolonging  AB. 

Ex.  491.  Through  one  of  the  points  of  intersection  of  two  circumfer- 
ences a  diameter  of  each  circle  is  drawn.  Prove  that  the  line  joining  the 
ends  of  the  diameters  passes  through  the  other  point  of  intersection. 


BOOK  II  149 


PROPOSITION  XXI.     PROBLEM 

373.   To  construct  a  tangent  to  a  circle  from  a  point  out- 
side. 


Given   circle  0  and  point  P  outside  the  circle. 
To  construct   a  tangent  from  P  to  circle  0. 

I.    Construction 

1.  Draw  PO. 

2.  With  Q,  the  mid-point  of  PO,  as  center,  and  with  QO  as 
radius,  construct  a  circumference  intersecting  the  circumference 
of  circle  0  in  points  T  and  F. 

3.  Draw  PTand  PF. 

4.  PT  and  PFare  tangents  from  P  to  circle  0. 

II.    The  proof  and  discussion  are  left  as  an  exercise  for  the 
student. 

HINT.     Draw  OT  and  OF  and  apply  §  367. 


Ex.  492.  Circumscribe  an  isosceles  triangle  about  a  given  circle,  the 
base  of  the  isosceles  triangle  being  equal  to  a  given  line.  What  restric- 
tion is  there  on  the  length  of  the  base  ? 

Ex.  493.  Circumscribe  a  right  triangle  about  a  given  circle,  one  arm 
of  the  triangle  being  equal  to  a  given  line.  What  is  the  least  length 
possible  for  the  given  line,  as  compared  with  the  diameter  of  the  circle  ? 

Ex.  494.  If  a  circumference  M  passes  through  the  center  of  a  circle 
B,  the  tangents  to  B  at  the  points  of  intersection  of  the  circles  intersect 
on  circumference  M. 

Ex.  495.  Circumscribe  an  isosceles  triangle  about  a  circle,  the  altitude 
upon  the  base  of  the  triangle  being  given. 


150  PLANE   GEOMETRY 

Ex.  496.     Construct  a  common  external  tangent  to  two  given  circles. 

r' X 

"  v 


M 


Given  circles  M TE  and  NV8. 

To  construct  a  common  external  tangent  to  circles  MTE  and  N  VS. 

I.    Construction 

1.  Draw  the  line  of  centers  OQ. 

2.  Suppose  radius  OM  >  radius   QN.     Then,  with  0  as  center  and 
with  OL  =  OM—  $xVas  radius,  construct  circle  LPK. 

3.  Construct  tangent  QP  from  point  Q  to  circle  LPK.     §  373. 

4.  Draw  OP  and  prolong  it  to  meet  circumference  M TE  at  T. 

5.  Draw  QV\\  OT.     §  188. 

6.  Draw  TV. 

7.  TV  is  tangent  to  circles  M  TE  and  NVS. 

II.    The  proof  and  discussion  are  left  as  an  exercise  for  the  student. 


Ex.  497.     Construct  a  second  common  external  tangent  to  circles  M  TE 
and      NVS     by      same 
method.       How    is    the 
method  modified    if   the 
two  circles  are  equal  ? 

Ex.  498.  Construct  a 
common  internal  tangent 
to  two  circles. 

HINT.  Follow  steps 
of  Ex.  490  except  step  2. 
Make  OL  =  OM  +  QN. 

Ex.  499.     By  moving 

Q  toward  O  in  the  preceding  figure,  show  when  there  are  four  common 
tangents  ;  when  only  three  ;  when  only  two  ;  when  only  one  ;  when  none. 


BOOK  II 


151 


PROPOSITION  XXII.     PROBLEM 

374.    With  a  given  line  as  chord,  to  construct  a  segment 
of  a  circle  capable  of  containing  a  given  angle. 


B 


Given  line  AB  and  Z  M. 

To  construct,  with  AB  as 
chord,  a  segment  of  a  circle 
capable  of  containing  Z  M. 

I.    Construction 

1.  Construct  an  Z,  as  /.CDE,  equal  to  the  given  Z  M.     §  125. 

2.  With  H,  any  convenient  point  on  DE,  as  center  and  with 
AB  as  radius,  describe  an  arc  cutting  DC  at  some  point,  as  K. 

3.  Draw  HK. 

4.  Circumscribe  a  circle  about  A  KDH.     §  323. 

5.  Segment  KDH  is  the  segment  capable  of  containing  Z  Jf. 
II.    The  proof -and  discussion  are  left  to  the  student. 

375.  Questions.     Without  moving  AB,  can  you  construct  a  A  with 
AB  as  base  and  a  vertex  Z  =  Z  M?     What  is  the  sum  of  the  base  A  ? 

376.  Cor.     The  locus  of  the  vertices  of  all  triangles  hav- 
ing a  given  base  and  a  given  angle  at  their  vertices  is  the 
arc  ivhich  forms,  with  the  given  base,  a  segment  capable 
of  containing  the  given  angle. 


Ex.  500.  On  a  given  line  construct  a  segment  that  shall  contain  an 
angle  of  105°  ;  of  135°. 

Ex.  501.  Find  the  locus  of  the  vertices  of  all  triangles  having  a  com- 
mon base  2  inches  long  and  having  their  vertex  angles  equal  to  60°. 

Ex.  502.     Construct  a  triangle,  having  given  6,  7i6,  and  B. 


152 


PLANE   GEOMETRY 


PROPOSITION  XXIII.     THEOREM 

377.  An  angle  formed  by  two  chords  which  intersect 
within  a  circle  is  measured  by  one  half  the  sum  of  the  arc 
intercepted  between  its  sides  and  the  arc  intercepted  be- 
tween the  sides  of  its  vertical  angle. 

r B 


Given   two  chords  AB  and  CD,  intersecting  at  E. 
To  prove    that  Z  1  oc  ^  (#/)  -j-  Jic). 


ARGUMENT 


Draw  AD. 


1. 
2. 

3.  Z  2  oc  i.  BD. 

4.  Z  3  oc  |  AC. 

5.  .-.  Z  2  +  ^  3  « 

6.  .-.  Z  1  oc  i.  (£D  4. 


+ 


Q.E.D. 


REASONS 
§  54,  15. 
§215. 
§365. 
§365. 
§362,6. 
§309. 


Ex.  503.  One  angle  formed  by  two  intersecting  chords  intercepts  an 
arc  of  40°.  Its  vertical  angle  intercepts  an  arc  of  60°.  How  large  is  the 
angle  ? 

Ex.  504.  If  an  angle  of  two  intersecting  chords  is  40°  and  its  inter- 
cepted arc  is  30°,  how  large  is  the  opposite  arc  ? 

Ex.  505.  If  two  chords  intersect  at  right  angles  within  a  circumfer- 
ence, the  sum  of  two  opposite  intercepted  arcs  is  equal  to  a  semicircum- 
ference. 

Ex.  506.  If  M  is  the  center  of  a  circle  inscribed  in  triangle  ABC  and 
if  AM  is  prolonged  to  meet  the  circumference  of  the  circumscribed  circle 
at  D,  prove  that  BD  =  DM  =  DC. 


BOOK  II 


153 


PROPOSITION  XXIV.     THEOREM 

378.   An  angle  formed  by  a  tangent  and  a  chord  is 
measured  by  one  half  its  intercepted  arc. 


B 


E 


Given  Z  ABC  formed  by  tangent  AB  and  chord  BC. 
To  prove    that  Z  AB C  9?  1  BC. 


ARGUMENT 

1.  Draw  diameter  BD. 

2.  Z.ABD  is  a  rt.  Z. 

3.  .-.  Z.ABD  «;  i  a  semicircumference ;  i.e. 

Z  ^5D  QC  i  arc  #<7£. 

4.  Z  CBD  QC  i.  cz). 

5.  .'.  /-ABD  —  Z  C/?Z>  Q?  -J-  (arc  .BCD  —  CD). 

.1    £&.  Q.E.D. 


REASONS 

1.  §  64,  15. 

2.  §313. 

3.  §297, 

4.  §365. 

5.  §  362,  6. 

6.  §309. 


Ex.  507.  In  the  figure  of  §  378,  if  arc  BC  =  100°,  find  the  number 
of  degrees  in  angle  ABC ;  in  angle  CBD  ;  in  angle  CBE. 

Ex.  508.  If  tangents  are  drawn  at  the  extremities  of  a  chord  which 
subtends  an  arc  of  120°,  what  kind  of  triangle  is  formed  ? 

Ex.  509.  If  a  tangent  is  drawn  to  a  circle  at  the  extremity  of  a 
chord,  the  mid-point  of  the  subtended  arc  is  equidistant  from  the  chord 
and  the  tangent. 

Ex.  510.     Solve  Prop.  XXII  by  means  of  Prop.  XXIV. 
HINT.     Observe  that  in  the  figure  for  Prop.  XXIV  any  angle  inscribed 
in  segment  BDC  would  be  equal  to  angle  ABC. 


« 
154 


PLANE   GEOMETRY 


PROPOSITION  XXV.     THEOREM 

379.  <An  angle  formed  by  two  secants  intersecting  out- 
side of  a  circumference,  an  angle  formed  by  a  secant  and 
a  tangent,  and  an  angle  formed  by  two  tangents  are  each 
measured  by  one  half  the  difference  of  the  intercepted  arcs. 


FIG.  2.  FIG.  3. 

I.    Aii  angle  formed  by  two  secants  (Fig.  1). 
Given     two  secants  BA  and  BC,  forming  Z  1. 

To  prove     that  Z  1  &  1  (AC  —  DE). 

t 
ARGUMENT 

Draw  CD. 

Z  1  +  Z.  2  =  Z  3. 

.--.  Z  1  =  Z  3  -  Z  2. 

Z3  K  %AC,  and  Z2  *  %  DE. 

.-.  Z3-Z2oc|(^_^). 

/.  Z  1  oc  i  (^  -  DE).  Q.E.D. 


REASONS 

1.  §  54,  15. 

2.  §  215. 

3.  §  54,  3. 

4.  §  365. 

5.  §362,6. 

6.  §  309. 


II.  An  angle  formed  by  a  secant  and  a  tangent  (Fig.  2). 

III.  An  angle  formed  by  two  tangents  (Fig.  3). 
The  proofs  of  II  and  III  are  left  to  the  student. 

380.  Note.  In  the  preceding  theorems  the  vertex  of  the  angle 
may  be  :  (1)  within  the  circle  ;  (2)  on  the  circumference  ;  (3)  outside 
the  circle.  _ 

Ex.  511.  Tell  how  to  measure  an  angle  having  its  vertex  in  each  of 
the  three  possible  positions  with  regard  to  the  circumference. 


BOOK  II 


« 
155 


PROPOSITION  XXVI.     THEOREM 

381.   Parallel  lines  intercept  equal  arcs  on  a  circum- 
ference.  A  H  B  H 


FIG.  1.  FIG.  2. 

I.   If  the  two  II  lines  are  secants  or  chords  (Fig.  1). 
Given     II  chords  AB  arid  CD,  intercepting  arcs  AC  and  BD. 
To  prove     AC  =  BD. 

REASONS 

1.  §  54,  15. 

2.  §  189. 


ARGUMENT 

1.  Draw  CB. 

2.  Z1  =  Z2. 

4.  .'.  ±AC=±BD. 

5.  .'.  AC  =  BD. 


Q.E.D. 


3.  §  365. 

4.  §  362,  a. 

5.  §  54,  7  a. 


II.  If  one  of  the  II  lines  is  a  secant  and  the  other  a  tangent 
(Fig.  2). 

III.  If  the  two  II  lines  are  tangents  (Fig.  3). 

The  proofs  of  II  and  III  are  left  as  exercises  for  the 
student. 

Ex.  512.  In  Fig.  1,  Prop.  XXV,  if  angle  1  equals  42°  and  arc  DE 
equals  40°,  how  many  degrees  are  there  in  angle  3  ?  in  arc  AC? 

Ex.  513.  In  Fig.  2,  if  arc  DO  equals  60°  and  angle  3  equals  100°,  find 
the  number  of  degrees  in  angle  1. 

Ex.  514.     In  Fig.  3,  if  angle  1  equals  65°,  find  angle  2  and  angle  3. 

Ex.  515.  An  angle  formed  by  two  tangents  is  measured  by  180° 
minus  the  intercepted  arc. 

Ex.  516.  If  two  tangents  to  a  circle  meet  at  an  angle  of  40°,  how 
many  degrees  of  arc  do  they  intercept  ? 


156  PLANE   GEOMETRY 

Ex.  517.     Is  the  converse  of  Prop.  XXVI  true  ?     Prove  your  answer. 

Ex.  518.  If  two  sides  of  an  inscribed  quadrilateral  are  parallel,  the 
other  two  sides  are  equal. 

Ex.  519.     Is  the  converse  of  Ex.  518  true  ?    Prove  your  answer. 

Ex.  520.  If.  through  the  point  where  the  bisector  of  an  inscribed 
angle  cuts  the  circumference,  a  chord  is  drawn  parallel  to  one  side  of  the 
angle,  this  chord  will  equal  the  other  side  of  the  angle. 

Ex.  521.  If  through  the  points  of  intersection  of  two  .circumferences 
parallels  are  drawn  terminating  in  the  circumferences,  these  parallels  will 
be  equal. 

Ex.  522.     Prove  that  a  trapezoid  inscribed  in  a  circle  is  isosceles. 

Ex.  523.  If  two  pairs  of  sides  of  an  inscribed  hexagon  are  parallel, 
the  other  two  sides  are  equal. 

Ex.  524.  If  the  sides  AB  and  BC  of  an  inscribed  quadrilateral  ABCD 
subtend  arcs  of  60°  and  130°,  respectively,  and  if  angle  AED,  formed  by 
the  diagonals  AC  and  BD  intersecting  at  E,  is  76°,  how  many  degrees  are 
there  in  arcs  AD  and  DC?  how  many  degrees  in  each  angle  of  the  quad- 
rilateral ? 

MISCELLANEOUS   EXERCISES 

Ex.  525.  Equal  chords  of  a  circle  whose  center  is  C  intersect  at  E. 
Prove  that  CE  bisects  the  angle  formed  by  the  chords. 

Ex.  526.  A  common  tangent  to  two  unequal  circles  intersects  their 
line  of  centers  at  a  point  P  \  from  P  a  second  tangent  is  drawn  to  one  of 
the  circles.  Prove  that  it  is  also  tangent  to  the  other. 

Ex.  527.  Through  one  of  the  points  of  intersection  of  two  circum- 
ferences draw  a  chord  of  one  that  shall  be  bisected  by  the  other 
circumference. 

Ex.  528.  The  angle  ABC  is  any  inscribed  angle  in  a  given  segment 
of  a  circle  ;  AC  is  prolonged  to  P,  making  CP  equal  to  CB.  Find  the  locus 
of  P. 

Ex.  529.  Given  two  points  P  and  $,  and  a  straight  line  through  Q. 
Find  the  locus  of  the  foot  of  the  perpendicular  from  Pto  the  given  line,  as 
the  latter  revolves  around  Q. 

Ex.  530.  If  two  circles  touch  each  other  and  a  line  is  drawn  through 
the  point  of  contact  and  terminated  by  the  circumferences,  the  tangents 
at  its  ends  are  parallel. 

Ex.  531.  If  two  circles  touch  each  other  and  two  lines  are  drawn 
through  the  point  of  contact  terminated  by  the  circumferences,  the  chords 
joining  the  ends  of  these  Hues  are  parallel. 


BOOK  II  15T 

Ex.  532.  If  one  arm  of  a  right  triangle  is  the  diameter  of  a  circle,  the 
tangent  at  the  point  where  the  circumference  cuts  the  hypotenuse  bisects 
the  other  arm. 

Ex.  533.  Two  fixed  circles  touch  each  other  externally  and  a  circle 
of  variable  radius  touches  both  externally.  Show  that  the  difference  of 
the  distances  from  the  center  of  the  variable  circle  to  the  centers  of  the 
fixed  circles  is  constant. 

Ex.  534.  If  two  circles  are  tangent,  externally,  their  common  internal 
tangent  bisects  their  common  external  tangent. 

Ex.  535.  If  two  circles  are  tangent  externally  and  if  their  common 
external  tangent  is  drawn,  lines  drawn  from  the  point  of  contact  of  the 
circles  to  the  points  of  contact  of  the  external  tangent  are  perpendicular 
to  each  other. 

Ex.  536.  The  two  common  external  tangents  to  two  circles  meet  their 
line  of  centers  at  a  common  point.  Also  the  two  common  internal  tan- 
gents meet  the  line  of  centers  at  a  common  point. 

Ex.  537.  Two  circles  whose  radii  are  17  and  10  inches,  respectively, 
are  tangent  externally.  How  long  is  the  line  joining  their  centers  ?  how 
long  if  the  same  circles  are  tangent  internally  ? 

Ex.  538.  If  a  right  angle  at  the  center  of  a  circle  is  trisected,  is  the 
intercepted  arc  also  trisected  ?  Is  the  chord  which  subtends  the  arc  tri- 
sected ? 

Ex.  539.  Draw  a  line  intersecting  two  given  circumferences  in  such 
a  way  that  the  chords  intercepted  by  the  two  circumferences  shall  equal 
two  given  lines.  What  restriction  is  there  on  the  lengths  of  the  given 
lines  ? 

Ex.  540.  Construct  a  triangle,  given  its  base,  the  vertex  angle,  and 
the  median  to  the  base.  Under  what  conditions  will  there  be  no  solution  ? 

Ex.  541.  In  the  same  circle,  or  in  equal  circles,  two  inscribed  triangles 
are  equal,  if  two  sides  of  one  are  equal  respectively  to  two  sides  of  the 
other. 

Ex.  542.  If  through  the  points  of  intersection  of  two  circumferences 
two  lines  are  drawn  terminating  in  the  circumferences,  the  chords  which 
join  their  extremities  are  parallel. 

Ex.  543.  The  tangents  drawn  through  the  vertices  of  an  inscribed 
rectangle,  which  is  not  a  square,  form  a  rhombus. 

Ex.  544.  The  line  joining  the  center  of  the  square  described  upon  the 
hypotenuse  of  a  right  triangle,  to  the  vertex  of  the  right  angle,  bisects  the 
right  angle. 


158 


PLANE   GEOMETRY 


Ex.  545.  If  two  common  external  tangents  or  two  common  internal 
tangents  are  drawn  to  two  circles,  the  segments  of  these  tangents  inter- 
cepted between  the  points  of  contact  are  equal. 

Ex.  546.  Through  two  given  points  draw  two  parallel  lines  at  a  given 
distance  apart. 

Ex.  547.  In  a  given  circle  inscribe  a  chord  of  given  length  which  pro- 
longed shall  be  tangent  to  another  given  circle. 

Ex.  548.  Find  the  locus  of  the  middle  point  of  a  chord  drawn  from  a 
given  point  in  a  given  circumference. 

Ex.  549.  The  locus  of  the  intersections  of  the  altitudes  of  triangles 
having  a  given  base  and  a  given  angle  at  the  vertex  is  the  arc  which 
forms  with  the  base  a  segment  capable  of  containing  an  angle  equal  to 
the  supplement  of  the  given  angle  at  the  vertex. 

HINT.  Let  ABC  be  one  of  the  &  and  0  the 
intersection  of  the  altitude's.  In  quadrilateral 
FOEC,  /.EOF  is  the  supplement  of  /.FOE. 
.-.  /.  EOF  is  the  supplement  of  /.BOA. 

Ex.  550.  In  a  circle,  prove  that  any  chord 
which  bisects  a  radius  at  right  angles  subtends  an 
angle  of  120°  at  the  center. 

Ex.  551.  Construct  an  equilateral  triangle,  having  given  the  radius 
of  the  inscribed  circle. 

Ex.  552.  The  two  circles  described  upon  two  sides  of  a  triangle  as 
diameters  intersect  upon  the  third  side. 

Ex.  553.  All  triangles 
circumscribed  about  the 
same  circle  and  mutually 
equiangular  are  equal. 

Ex.  554.  If  two  cir- 
cumferences meet  on  their 
line  of  centers,  the  circles 


R 


FIG.  1. 


FIG.  2. 


are  tangent  to  each  other. 


OUTLINE  OF  PROOF 


M  between  0  and  Q,  Fig.  1. 

QP+PQ>OQ. 

OP=  OM. 

.'.PQ>MQ. 

/.  P  is  not  on  circumference  S. 


II.   3/not  between  O  and  Q,  Fig.  2. 
OQ+  QP>OP. 
.'.  OQ+QP>OM. 
.:QP>  QM. 
.'.  P  is  not  on  circumference  S. 


BOOK   II  159 

Ex.  555.  If  two  circumferences  intersect,  neither  point  of  intersection 
is  on  their  line  of  centers. 

Ex.  556.  In  any  right  triangle  ABC,  right-angled  at  C,  the  radius 
of  the  inscribed  circle  equals  \(a  +  b  —  c)  and  the  radius  of  the  escribed 
circle  tangent  to  c  equals  |(«  +  b  +  c). 

Ex.   557.     In   the    accom-       .    ? B  P 

panying  figure  a,  &,  c,  are  the 
sides  of  triangle  ABC.    Prove  : 

1.  a  =  TP; 

2.  AP= 

3.  TB  =  ±( 

Ex.  558.     Trisect  a  quadrant ;  a  semicircumference  ;  a  circumference. 

Ex.  559.  Describe  circles  about  the  vertices  of  a  given  triangle  as 
centers,  so  that  each  shall  touch  the  other  two. 

Ex.  560.  Construct  within  a  given  circle  three  equal  circles,  so  that 
each  shall  touch  the  other  two  and  also  the  given  circle. 

Construct  a  triangle,  having  given  : 
Ex.  561.    ha,  hc,  C. 

Ex.  562.     A,  B,  and  JR,  the  radius  of  the  circumscribed  circle. 
Ex.  563.     a,  B,  It. 

Ex.  564.     C  and  the  segments  of  c  made  by  tc. 
Ex.  565.     C  and  the  segments  of  c  made  by  hc. 

Ex.  566.  r,  the  radius  of  the  inscribed  circle,  and  the  segments  of 
c  made  by  tc. 

Construct  a  right  triangle  ABC,  right-angled  at  (7,  having  given : 

Ex.  567.     c,  he. 

Ex.  568.     c  and  one  segment  of  c  made  by  hc. 

Ex.  569.     The  segments  of  c  made  by  hc. 

Ex.  570.     The  segments  of  c  made  by  tc. 

Ex.  571.     c  and  a  line  I,  in  which  the  vertex  of  C  must  lie. 

Ex.  572.     c  and  the  perpendicular  from  vertex  C  to  a  line  I. 

Ex.  573.     c  and  the  distance  from  C  to  a  point  P. 

Construct  a  square,  having  given  : 

Ex.  574.     The  perimeter. 

Ex.  575.     A  diagonal. 

Ex.  576.     The  sum  of  a  diagonal  and  a  side. 


160  PLANE   GEOMETRY 

Construct  a  rectangle,  having  given  : 

Ex.  577.     Two  non-parallel  sides. 

Ex.  578.     A  side  and  a  diagonal. 

Ex.  579.     The  perimeter  and  a  diagonal. 

Ex.  580.     A  diagonal  and  an  angle  formed  by  the  diagonals. 

Ex.  581.     A  side  and  an  angle  formed  by  the  diagonals. 

Ex.  582.     The  perimeter  and  an  angle  formed  by  the  diagonals. 

Construct  a  rhombus,  having  given : 

Ex.  583.     A  side  and  a  diagonal. 

Ex.  584.     The  perimeter  and  a  diagonal. 

Ex.  585.     One  angle  and  a  diagonal. 

Ex.  586.     The  two  diagonals. 

Ex.  587.     A  side  and  the  sum  of  the  diagonals. 

Ex.  588.     A  side  and  the  difference  of  the  diagonals. 

Construct  a  parallelogram,  having  given : 

Ex.  589.     Two  non-parallel  sides  and  an  angle. 

Ex.  590.     Two  non-parallel  sides  and  a  diagonal. 

Ex.  591.     One  side  and  the  two  diagonals. 

Ex.  592.     The  diagonals  and  an  angle  formed  by  them. 

Construct  an  isosceles  trapezoid,  having  given : 

Ex.  593.     The  bases  and  a  diagonal. 

Ex.  594.     The  longer  base,  the  altitude,  and  one  of  the  equal  sides. 

Ex.  595.     The  shorter  base,  the  altitude,  and  one  of  the  equal  sides. 

Ex.  596.     Two  sides  and  their  included  angle. 

Construct  a  trapezoid,  having  given : 

Ex.  597.     The  bases  and  the  angles  adjacent  to  one  base. 

Ex.  598.     The  bases,  the  altitude,  and  an  angle. 

Ex.  599.     One  base,  the  two  diagonals,  and  their  included  angle. 

Ex.  600.     The  bases,  a  diagonal,  and  the  angle  between  the  diagonals. 

Construct  a  circle  which  shall : 

Ex.  601.     Touch  a  given  circle  at  P  and  pass  through  a  given  point  Q. 

Ex.  602.     Touch  a  given  line  I  at  P. 

Ex.  603.     Touch  three  given  lines  two  of  which  are  parallel. 

Ex.  604.     Touch  a  given  line  I  at  P  and  also  touch  another  line  m. 

Ex.  605.     Have  its  center  in  line  Z,  cut  I  at  P,  and  touch  a  circle  K. 


BOOK   III 

PROPORTION   AND    SIMILAR   FIGURES 

382.  Def.    A  proportion  is  the  expression  of  the  equality  of 
two  ratios. 

EXAMPLE.     If  the  ratio  -  is  equal  to  the  ratio  -,  then  the  equation 
b  d 

-  =  -  is  a  proportion.    This  proportion  may  also  be  written  a  :  b  =  c  :  d  or 
6      d 

a  :  b  : :  c  :  d,  and  is  read  a  is  to  b  as  c  is  to  d. 

383.  Def.    The  four  numbers  a,  b,  c,  d  are  called  the  terms 
of  the  proportion. 

384.  Def.    The  first  term  of  a  ratio  is  called  its  antecedent 
and  the  second  term  its  consequent ;  therefore  : 

The  first  and  third  terms  of  a  proportion  are  called  ante- 
cedents, and  the  second  and  fourth  terms,  consequents. 

385.  Def.    The  second  and  third  terms  of  a  proportion  are 
called  its  means,  and  the  first  and  fourth  terms,  its  extremes. 

386.  Def.    If  the  two  means  of  a  proportion  are  equal,  this 
common  mean  is  called  the  mean  proportional  between  the  two 
extremes,  and  the  last  term  of  the  proportion  is  called  the  third 
proportional  to  the  first  and  second  terms  taken  in  order ;  thus, 
in  the  proportion  a  :  b  =  b  :  c,  b  is  the  mean  proportional  between 
a  and  c,  and  c  is  the  third  proportional  to  a  and  b. 

387.  Def.    The  fourth  proportional  to  three  given  numbers 
is  the  fourth  term  of  a  proportion  the  first   three   terms    of 
which  are  the  three  given  numbers  taken   in  order ;  thus,   if 
a :  b  =  c :  d,  d  is  called  the  fourth  proportional  to  a,  b,  and  c. 

161 


162  PLANE   GEOMETRY 

PROPOSITION  I.     THEOREM 

388.     If  four  numbers  are  in  proportion,  the  product 
of  the  extremes  is  equal  to  the  product  of  the  means. 
Given    a  :  b  =  c  :  d. 
To  prove    ad  =  be. 


1. 


ARGUMENT 

a  __  G 

b~d' 


2.  bd  =  bd. 

3.  .-.ad  =  bc.  Q.E.D. 


REASONS 

1.  By  hyp.* 

2.  By  iden. 

3.  §  54,  la. 


389.  Note.     The  student  should  observe  that  the  process  used  here 
is  merely  the  algebraic  "clearing  of  fractions,"  and  that  as  fractional 
equations  in  algebra  are  usually  simplified  by  this  process,  so,  also,  pro- 
portions may  be  simplified  by  placing  the  product  of  the  means  equal  to 
the  product  of  the  extremes. 

390.  Cor.  I.    The  mean  proportional  between  two  num- 
bers is  equal  to  the  square  root  of  their  product. 

391.  Cor.  II.    If  two  proportions  have  any  three  terms 
of  one  equal  respectively  to  the  three  corresponding  terms 
of  the  other,  then  the  remaining  term  of  the  first  is  equal 
to  the  remaining  term  of  tfa  second. 


Ex.  606.  Given  the  equation  m  :  r  =  d  :  c  :  solve  (1 )  for  d,  (2)  for  r, 
(3)  for  m,  (4)  for  c. 

Ex.  607.  Find  the  fourth  proportional  to  4,  6,  and  10 ;  to  4, 10,  and  0; 
to  10,  6,  and  4. 

Ex.  608.  Find  the  mean  proportional  between  9  and  144 ;  between 
144  and  9. 

Ex.  609.    Find  the  third  proportional  to  f  and  f  ;  to  f  and  $. 


392.      Questions.     What  rearrangement  of  numbers  can  be  made 
in  Ex.  (507  without  affecting  the  required  term  ?  in  Ex.  008  ?  in  Ex.  009  ? 
Ex.  610.     Find  the  third  proportional  to  a2  —  &2  and  a  —  b. 
Ex.  611.     Find  the  fourth  proportional  to  a2  —  62,  a  —  ft,  and  a  +  b. 
*  See  §382  for  the  three  ways  of  writing  a  proportion. 


BOOK  III  163 

Ex.  612.  If  in  any  proportion  the  antecedents  are  equal,  then  the 
consequents  are  equal  and  conversely. 

Ex.  613.     If  a  :  b  =  c  :  d,  prove  that  ma  :  kb  =  me  :  kd. 
Ex.  614.     If  I  :  k  =  b  :  m,  prove  that  Ir  :  kr  =  be  :  me. 
Ex.  615.     If  x  :  y  =  b  :  c,  prove  that  dx  :  y  =  bd  :  c. 
Ex.  616.     If  x  :  y  =  b  :  c,  is  dx  :  y  =  6  :  cd  a  true  proportion  ? 

PROPOSITION  II.     THEOREM 
(Converse  of  Prop.  I) 

393  If  the  product  of  two  numbers  is  equal  to  the  prod- 
uct of  two  other  numbers,  either  pair  may  be  made  the 
means  and  the  other  pair  the  extremes  of  a  proportion. 

Given    ad  =  be. 

To  prove    a:b  =  c:d. 


ARGUMENT 

1.  ad  =  bc. 

2.  bd  =  bd. 

3.  /.-  =  -;  i.e.  a :  b  =  c :  d.      Q.E.D. 

6      d 


REASONS 

1.  By  hyp. 

2.  By  iden. 

3.  §  54,  8  a. 


The  proof  that  a  and  d  may  be  made  the  means  and  b  and  c 
the  extremes  is  left  as  an  exercise  for  the  student. 

394.  Note.  The  pupil  should  observe  that  the  divisor  in  Arg.  2 
above  must  be  chosen  so  as  to  give  the  desired  quotient  in  the  first  mem- 
ber* of  the  equation  :  thus,  if  hi  =  kf,  and  we  wish  to  prove  that-  =  -,  we 

must  divide  by  fl ;  then  M  =  &,  i.e.  \  =* 
fl      fl         J      I 


Ex.  617.     Given  pt  =  cr.     Prove  p  :  r  =  c  :  t ;  also,  c:p  =  t:r. 
Ex.  618.     From  the  equation  rs  =  Im,  derive  the  following  eight  pro- 
portions : 

r  :  I  =  m  :  s,  s:l  =  m:r,  l:r  =  s: »»,  m  :  r  =  s  :  I 

r  :  m  =  I  :  s,  s  :  m  —  I :  r,  I :  s  =  r  :  w,  m  :  s  =  r  :  I. 

Ex.  619.     Form  a  proportion  from  7  x  4  =  3  x  a  ;  from  ft  =  gb.     How 
can  the  proportions  obtained  be  verified  ? 


164 


PLANE   GEOMETRY 


Ex.  620.     Form  a  proportion  from  (a  +  c)(a  —  b)  =  de. 
Ex.  621.     Form  a  proportion  from  m'2  —  2  mil  +  n2  =  db. 
Ex.  622.     Form  a  proportion  from  c'2  +  2  cd  +  d'2  =  a  +  6. 
Ex.  623.     Form  a  proportion  from  (a  +  &)(«  —  6)  =4se, 
making  x  (1)  an  extreme  ;  (2)  a  mean. 

Ex.  624.     If7 x  +  By  :l2  =  2x+y  :  3,  find  the  ratio  x  : y. 


PROPOSITION  III.     THEOREM 

395.  If  four  numbers  are  in  proportion,  they  are  in 
proportion  by  inversion;  that  is,  the  second  term  is  to  the 
first  as  tlie  fourth  is  to  the  third. 

Given     a  :  b  =  c :  d. 
To  prove     b:a  =  d:c. 


ARGUMENT 
a  :  b  =  c :  d. 
.-.  ad  =  bc. 


Q.E.D. 


REASONS 

1.  By  hyp. 

2.  §  388. 

3.  §  393. 


PROPOSITION  IV.     THEOREM 

396.   Tf  four  numbers  are  in  proportion,  they  are  in 
proportion  by  alternation;  that  is,   the  first  term  is  to 
third  as  the  second  is  to  the  fourth. 


Given     a:b  =  c:d. 
To  prove     a :  c  =  b :  d. 


1. 
2. 
3. 


ARGUMENT 
a  :  b  =  c  :  d. 

.•.  a:  c  =  b  :d. 


Q.E.D. 


REASONS 

1.  By  hyp. 

2.  §388. 

3.  §  393. 


Ex.  625.     If  x  : J/  =  y  :  |,  what  is  the  value  of  the  ratio  x  :  y  ? 
Ex.  626.     Transform  a  :  x  =  4  :  3  so  that  x  shall  occupy  in  turn  every 
place  in  the  proportion. 


BOOK  III  165 

397.  Many  transformations  may  be  easily  brought  about  by 
the  following  method : 

(1)  Reduce  the  conclusion   to   an  equation  in  its  simplest 
form  (§  388),  then  from  this  derive  the  hypothesis. 

(2)  Begin  with  the  hypothesis  and  reverse  the  steps  of  (1). 
This  method  is  illustrated  in  the  analysis  of  Prop.  V. 

PROPOSITION  V.     THEOREM 

398.  If  four  numbers  are  in   proportion,  they  are   in 
proportion  by  composition;  that  is,  the  sum  of  the  first 
two  terms  is  to   the  first   (or  second)  term  as  the  sum 
of  the  last  two  terms  is  to  the  third  (or  fourth)  term. 

Given     a:b  =  c:d. 
To  prove :      (a)  a  +  &:o  =  c4-d:c; 
(b)  a  +  b :  b  =  c  +  d :  d. 

I.    Analysis 

(1)  The  conclusions  required  above,  when  reduced  to  equa- 
tions in  their  simplest  forms,  are  as  follows: 

(a)  (b) 

1.  ac -\-ad  =  ac  +  bc.  1.  bc  +  bd  =  ad  +  bd. 

2.  Whence  ad  =  be.  2.  Whence  be  =  ad. 

3.  .:  a:b  =  c:d.  3.  .:  a:b  =  c:d. 

(2)  Now  begin  with  the  hypothesis  and  reverse  the  steps. 


II.    Proof 

(d)  ARGUMENT 

1.  a:b  =  c:d. 

2.  .-.  ad  =  be. 

3.  ac  =  ac. 

4.  .•.ac+ad=ac-)-bc] 
i.e.  a(c  +  d)  =  c(a  +  6). 

5.  .-.a  +b:  a  =  c  +  d :  e.          Q.E.D. 


REASONS 

1.  By  hyp. 

2.  §  388. 

3.  By  iden. 

4.  §  54,  2. 


5.   §393. 
(b)  The  proof  of  (b)  is  left  as  an  exercise  for  the  student. 


166  PLANE   GEOMETRY 

Ex.  627.     If  -  =  -,  fin 

2/5  x  y 

PROPOSITION   VI.     THEOREM 

399.  If  four  numbers  are  in  proportion,  they  are 
in  proportion  by  division;  that  is,  the  difference  of  the 
first  two  terms  is  to  the  first  (or  second)  term  as  the  differ- 
ence of  the  last  two  terms  is  to  the  third  (or  fourth)  term. 

Given     a :  b  =  c :  d. 

To  prove  :     (a)  a  —  b:a  =  c  —  d :  c ; 

(6)  a  —  b :  b  ==  c  —  d :  d. 

I.   The  analysis  is  left  as  an  exercise  for  the  student. 


II.    Proof 

(a)  ARGUMENT 

1.  a:  b  =  c:  d. 

2.  .-.  ad  =  be. 

3.  ac  =  ac. 

4.  .-.  ac  —  ad  =  ac  —  be, 
i.e.  a(c  —  d)  =  c(a  —  5). 

5.  .'.  a  —  b:a  =  c  —  d:  c.  Q.E.D. 


REASONS 

1.  By  hyp. 

2.  §388. 

3.  By  iden. 

4.  §54,3. 


5.    §393. 
(b)  The  proof  of  (b)  is  left  as  an  exercise  for  the  student. 


Ex.   628.    H?=|,  find^rJ!;    — 

y     3  x  y 


PROPOSITION   VII.     THEOREM 

400.  If  four  numbers  are  in  proportion,  tlwy  are 
in  proportion  by  composition  and  division;  that  is,  the 
sum  of  the  first  two  terms  is  to  their  difference  as  the  sum 
of  the  last  two  terms  is  to  their  difference. 

Given     a  :  b  =  c :  d. 

To  prove     d-}-  b:  a  —  6  =  c  +  d:c—  d. 


BOOK   III 


167 


ARGUMENT 

1. 

a:b 

=  c:d. 

o 

a  -\-b 

c  +  d 

a 

c 

•-?        Anrl 

a-b 

c-d 

a 

c 

4. 

a  +  b 

c  +  d. 

REASONS 

1.  By  hyp. 

2.  §398. 

3.  §399. 

4.  §  54,  8  a. 


PROPOSITION  VIII.     THEOREM 

401.  In  a  series  of  equal  ratios  the  sum  of  any  number 
of  antecedents  is  to  the  sum  of  tfo  corresponding  conse- 
quents as  any  antecedent  is  to  its  consequent. 

Given     a  :  b  =  c :  d  =  e  :/  =  g :  h. 

To  prove     a  +  c  -f-  e  +  g  :b  -\-  d  -f- / -f  h  =  a:b. 

I.    Analysis 

Simplifying  the  conclusion  above,  we  have : 

ab  -f-  be  -{-  be  -f-  bg  =  ab  +  ad  -f-  of  +  ah. 

The  terms  required  for  the  first  member  of  this  equation, 
and  their  equivalents  for  the  second  member,  may  be  obtained 
from  the  hypothesis. 

II.    Proof 

ARGUMENT  REASONS 

ab  =  ab.  1.  By  iden. 

bc  =  ad.  2.  §388. 

be  =  af.  3.  §388. 

bg  =  ah.  4.  §388.. 

5.  .-.  ab  +  be  -f  be  +  bg  =  ab  +  ad  +af+  ah ;  5.  §  54,  2. 
i.e.  b  (a  +c  -h  e  +  r/)  =  a  (b  +  d  -f-/+  7t). 

6.  .-.  a+  c  +  e  +  0:&  +  d+/+7i  =  a:6.  6.  §393. 

Q.E.D. 


168 


PLANE   GEOMETRY 


Ex.  629.     With  the  hypothesis  of  Prop.  VIII,  prove 
a  +  c+e:b  +  d+f=g:h. 

Ex.  630.     If  -  =  -  =  -  =  £,  prove  m~s  +  k~P  _ 
T      t     g      q  r  —  t  +  g  —  q 


Ex.  631.     If 


=  -,  find  -.       HINT.     Use  Prop.  VI. 


Ex.  632.     If  a  :  b  —  c  :  d,  show  that  b  —  a:  b  +  a  =  d  —  c  :  d  +  c. 

Ex.  633.  Given  the  proportion  a  :  b  =  11 :  6.  Write  the  proportions 
that  result  from  taking  the  terms  (1)  by  inversion  ;  (2)  by  alternation  ; 
(3)  by  composition  ;  (4)  by  division  ;  (5)  by  composition  and  division. 

PROPOSITION  IX.     THEOREM 

402.  The  products  of  the  corresponding  terms  of  any 
number  of  proportions  form  a  proportion. 

Given     a  :  b  =  c :  c7,  e:f=g:h,  i:j  =  k:l. 
To  prove     aei :  bfj  =  cgk :  dhl. 


1. 


ARGUMENT 
a      c 


/ 


2. 


i  _  cgk  t 
' 


.e.  aei  :       = 


Q.E.D. 


REASONS 
1.   By  hyp. 


2.    §54,  7  a. 


403.  Cor.  //  four  numbers  are  in  proportion,  equi- 
multiples of  the  first  two  and  equimultiples  of  the  last 
two  are  also  in  proportion. 

HINT.     Given     a  :  b  =  x  :  y. 

To  prove     am  :bm  =  nx:  ny. 


Ex.  634.     ltr:s  =  m:t,  is  fr  :  qs  =  qm  :ft ?     Prove  your  answer. 
Ex.   635.     If  a  :  b  —  3  :  4  and  x  :  y  =  8  : 0,  find  the  value  of  ax  :  by. 


BOOK  III 


169 


Ex.   636.     If  four  numbers  are  in  proportion,  equimultiples   of  the 
antecedents  and  equimultiples  of  the  consequents  are  also  in  proportion. 

Ex.   637.     If  r  :  s  =  t :  m,  prove  3a  +  2w:4s  =  3r  +  2«:4r. 

Ex.  638.     Iiw:x  =  y:z,  prove  ax  +  bz  :  ex  -\-  dz  =  aw  +  by  :  cw  +  dy. 

Ex.  639.     If  —  =  -  and  -  =  -  ,  prove  that  —  :  -  =  -  :  - ,  and  state  the 
r      v  s     w  n   s       I    w 

theorem  thus  derived. 


PROPOSITION  X.     THEOREM 

404.   If  four  numbers  are  in  proportion,  like  powers  of 
these  numbers  are  in  proportion,  and  so  also  are  like  roots. 

Given    a  :  b  —  c  :  d. 

To  prove  :    (a)  ap  :bp  =  cp  :  dp. 

(b)   Va  :  V&  =  Vc  :  - 


REASONS 
1.    By  hyp. 

=  <?:&.  2.    §54,13. 


ARGUMENT 

1. 

a_  C 

6     d' 

2. 

ap 

cp 
=  —  ;  i.e.  ap  :  < 

Vc 

3.   J 

Llso  — 

=  "77-  >  i-e-  v  t 

Q.E.D. 


3.    §  54,  13. 


Ex.640. 


s      q 


Vs 


3s 


Ex.  641.     If  ™  =P  =  r-  ,  prove  that 

n      q      s  n 


405.  Def.  A  continued  proportion  is  a  series  of  equal  ratios 
in  which  the  consequent  of  any  ratio  is  the  same  number  as 
the  antecedent  of  the  following  ratio  ;  thus, 

a  :  b  =  c  :  d  —  e  :  f=  g:h\s  merely  a  series  of  equal  ratios,  while 
a  :  6  =  6  :  c  =  c  :  d  =  d  :  €  is  not  only  a  series  of  equal  ratios  but 
a  continued  proportion  as  well. 


170  PLANE   GEOMETRY 

Ex.  642.     In  the  continued  proportion  a  :b  =  b  :  c  =  c:d  =  d:et  prove 
that  : 

£  =  «!.    «  =  £;  and  ®=2!. 

c      b2 '   d      63  e      6* 

Ex.  643.     If  x3  :  ?/3  =  8  :  27,  find  - . 

y 

Ex.644.     If  ^/m:l  =  Vn  :  16,  find  -. 


406.  Def.     The  segments  of  a  line  are  the  parts  into  which 
it  is  divided.     The  line  AE  is  divided  internally  at  C  if  this 
point  is  between  the  extremities  of  the  line.     The  segments 
into  which  it  is  divided  are  A  C  and  CB. 

D  A C B 

AB  is  divided  externally  at  D  if  this  point  is  on  the  prolonga- 
tion of  the  line.  The  segments  are  AD  and  DB. 

It  should  be  noted  that  in  either  case  the  point  of  division  is 
one  end  of  each  segment. 

407.  Def.    Two   straight  lines  are  divided  proportionally  if 
the  ratio  of  one  line  to  either  of  its  segments  is  equal  to  the 
ratio  of  the  other  line  to  its  corresponding  segment. 

408.  In  Prop.  XT,  II,  the  following  theorems  (Appendix, 
§§  586  and  591)  will  be  assumed : 

(a)  The  quotient  of  a  variable  by  a  constant  is  a  variable. 

(b)  The  limit  of  the  quotient  of  a  variable  by  a  constant  is  the 
limit  of  the  variable  divided  by  the  constant. 

Thus,  if  x  is  a  variable  and  k  a  constant : 

(1)  -  is  a  variable. 

K 

1*  ?/ 

(2)  If  the  limit  of  x  is  ?/,  then  the  limit  of  -  is  ^. 

/C          K 

Ex.  645.  Tn  the  figure  of  §  409,  name  the  segments  into  which  AB 
is  divided  by  D ;  the  segments  into  which  AD  is  divided  bv  B. 


BOOK  III 


171 


PROPOSITION  XI.     THEOREM 

409.   <A  straight  line  parallel  to  one  side  of  a  triangle 
divides  the  other  two  sides  proportionally. 


E 


FIG.  1. 


Given   A  ABC  with  line  DE  II  BC. 


AB      AC 
To  prove    —  =  — . 

AD      AE 


I.   If  AB  and  AD  are  commensurable  (Fig.  1). 


ARGUMENT 

1.  Let  AF  be  a  common  measure  of  AB 

and  AD,  and  suppose  that  AF  is  con- 
tained in  AB  r  times  and  in  AD  s 
times. 

2.  Then  ^  =-. 

^Z>      s 

3.  Through  the  several  points  of  division 

on  AB,  as  F,  G,  etc.,  draw  lines  ||  BC. 

4.  These  lines  are  ||  DE  and  to  each  other. 

5.  .•.  AC  is  divided  into  r  equal  parts  and 

AE  into  s  equal  parts. 

6.  .:^=r-. 

AE       s 

7.  ...   ^?  =  £^.  Q.E.D. 

AD       AE 


REASONS 
1.    §  335. 


2.  §  341. 

3.  §  179. 

4.  §  180. 

5.  §  244. 

6.  §  341. 

7.  §54,1, 


172  PLANE   GEOMETRY 

II.    If  AB  and  AD  are  incommensurable  (Fig.  2). 


1. 


6. 


E 

FIG.  2. 


KC 


ARGUMENT 

Let  m  be  a  measure  of  AD.  Apply  m 
as  a  measure  to  AB  as  many  times 
as  possible.  There  will  then  be  a 
remainder,  HB,  less  than  m. 

AH  and  AD  are  commensurable. 

Draw  HK\\  BC. 
AH  =  AK 

'  AD~  AE' 

Now  take  a  smaller  measure  of  AD. 
No  matter  how  small  a  measure  of 
AD  is  taken,  when  it  is  applied  as 
a  measure  to  AB,  the  remainder,  HB, 
will  be  smaller  than  the  measure 
taken. 

.-.  the  difference  between  AH  and  AB 
may  be  made  to  become  and  remain 
less  than  any  previously  assigned 
line,  however  small. 

.-.  AH  approaches  AB  as  a  limit. 

.»,  —  approaches  —  as  a  limit. 
AD  AD 


9.  Likewise  the  difference  between  AK 
and  AC  may  be  made  to  become  and 
remain  less  than  any  previously  as- 
signed line,  however  small, 


REASONS 
1.    §  339. 


2.  §337. 

3.  §  179. 

4.  §  409,  I. 

5.  §335. 


6.   Arg.  5. 


7.  §  349. 

8.  §  408,  b. 

9.  Arg.  6. 


BOOK  III  173 


ARGUMENT 


10.    .•.  AK  approaches  AC  as  a  limit. 


AK  A  C1 


11.   /.  -  -  approaches  -  -  as  a  limit. 


AE  AE 

Apr  A~K 

12.   But  -  -  is  always  equal  to  —  . 
AD  AE 

1Q          AB       AC 

10.     .*.  --  =  -  .  Q.E.D. 

AD       AE 


REASONS 

10.  §  349. 

11.  §408,6. 

12.  Arg.  4. 


13.    §355. 


410.  Cor.  A  straight  line  parallel  to  one  side  of  a  tri- 
angle divides  the  other  two  sides  into  segments  which  are 
proportional.  Thus,  in  the  figures  for  Prop.  XI,  AD :  DB 
=  AE :  EC. 

HINT.     Prove  by  using  division. 


Ex.  646.     Using  Fig.  2  of  Prop.  XI,  prove 

(1)  AB:AC=AD:AE;  (2)  AB  :  AC  =  DB  :  EC. 
Ex.  647.     In  the  diagram  at  the  right,  w,  n, 


and  r  are  parallel  to  each  other.     Prove  that  a  :  b     — \ 


•m 


=  c  :  d  ;  also  that  a  :  c  =  b  :  d.  \ 

Ex.  648.     If   two  sides   of   a  triangle   are  12  P\ 

inches  and  18  inches,  and  if  a  line  is  drawn  paral-  \ 

lei  to  the  third  side  and  cuts  off  3  inches  from  the 
vertex  on  the  12-inch  side,  into  what  segments  will  it  cut  the  18-inch  side  ? 

Ex.  649.  If  two  lines  are  cut  by  any  number  of  parallels,  the  two 
lines  are  divided  into  segments  which  are  proportional :  (a)  if  the  two 
lines  are  parallel ;  (6)  if  the  two  lines  are  oblique. 

Ex.  650.  Jf  through  the  point  of  intersection  of  the  medians  of  a 
triangle  a  line  is  drawn  parallel  to  any  side  of  the  triangle,  this  line 
divides  the  other  two  sides  in  the  ratio  of  2  to  1. 

Ex.  651.  A  line  can  be  divided  at  but  one  point  into  segments  which 
have  a  given  ratio  (measured  from  one  end). 

Ex.  652.  A  line  parallel  to  the  bases  of  a  trapezoid  divides  the  other 
two  sides  and  also  the  two  diagonals  proportionally. 

Ex.  653.  Apply  the  proof  of  Prop.  XI  to  the  case  in  which  the  par- 
allel to  the  base  cuts  the  sides  prolonged :  (a)  through  the  ends  of  the 
base ;  (&)  through  the  vertex. 


174  PLANE   GEOMETRY 


PROPOSITION  XII.     PROBLEM 

411.   To  construct  the  fourth  proportional  to  three  given 
lines. 


R  \M      b         \L        S 

Given   lines  a,  b,  and  c. 
To  construct   the  fourth  proportional  to  a,  b,  and  c. 

I.    Construction 

1.  From  any  point,  as  R,  draw  two  indefinite  lines  RS  and 
RT. 

2.  On  RS  lay  off  RM  =  a  and  ML  =  b. 

3.  On  RT  lay  off  RF  =  c. 

4.  Draw  MF. 

5.  Through  L  construct  LG  II  MF.     §  188. 

6.  FG  is  the  fourth  proportional  to  a,  b,  and  c. 

II.   The  proof  and  discussion  are  left  as  an  exercise  for  the 
student. 

412.  Question.     Could  the  segments  a,  ft,  and  c  be  laid  off  in  any 
other  order  ? 

413.  Cor.  I.     To  construct  the  third  proportional  to  two 
given  lines. 

414.  Cor.  II.     To  divide  a  given  line  into  segments  pro- 
portional to  two  or  more  given  lines. 


Ex.  654.     Divide  a  given  line  into  segments  in  the  ratio  of  3  to  5. 
Ex.  655.     Divide  a  given  line  into  segments  proportional  to  2,  3,  and  4. 


BOOK   III 


175 


Ex.  656.     Construct  two  lines,  given  their  sum  and  their  ratio. 
Ex.  657.     Construct  two  lines,  given  their  difference  and  their  ratio. 
Ex.  658.     If  a,  &,  and  c  are  three  given  lines,  construct  x  so  that : 
(a)  x:a  =  b:c;  (6)  x  =      - 


PROPOSITION  XIII.     THEOREM 
(Converse  of  Prop.  XI) 

415.   If  a  straight  line  divides  two  sides  of  a  triangle 
proportionally,  it  is  parallel  to  the  third  side. 


F  E 


C 


Given   A  ABC.  and  DE  so  drawn  that  —  =  — . 

AD       AE 
To  prove  DE  II  EG. 


ARGUMENT 

1.  DE  and  EC  are  either  II  or  not  II. 

2.  Suppose  that  DE  is  not  II  BC,  but  that 

some  other  line  through  D,  as  DF,  is  II 
BC. 

3rp-i          AE         AC> 
.    Inen  —  =  — . 
AD       AF 

4.  But    ^  =  ^. 

AD       AE 

5.  .'.          AF  =  AE. 

6.  This  is  impossible. 

7.  .*.  DE  II   BC.  Q.E.D. 


REASONS 

1.  §161,  a. 

2.  §179. 


3.  §409. 

4.  By  hyp. 

5.  §391. 

6.  §-54,12. 

7.  §  161,  b. 


416.  Cor.  If  a  straight  line  divides  two  sides  of  a  tri- 
angle into  segments  which  are  proportional;,  it  is  parallel 
to  the  third  side.  Thus,  if  AD :  DB  =  AE :  EC,  DE  is  II  BC. 


176  PLANE   GEOMETRY 

Ex.  659.  In  the  diagram  at  the  right,  if  AB 
=  15,  AC  -  12,  AD  =  10,  and  AE  =  8,  prove 
DE  parallel  to  BC. 

Ex.  660.     If  AB  =  60,  DB  =  15,  AE  =  28, 

and  EC  =  12,  is  DE  parallel  to  BC  ?    Prove. 

^1  J* 

Ex.  661.  If  DE  ||  BC,  AB  =  25,  DB  =  5, 
and  AC  =  20,  find  AE. 

Ex.  662.     If  DE  ||  BC,  AB=  80,  ^Z>  =  25,  and  EC  =  4,  find 


SIMILAR  POLYGONS 

417.  Def.     If  the  angles  of  one  polygon,  taken  in  order,  are 
equal  respectively  to  those  of  another,  taken  in  order,  the  poly- 
gons are  said  to  be  mutually  equiangular.     The  pairs  of  equal 
angles  in  the  two  polygons,  taken  in  order,  are  called  homolo- 
gous angles  of  the  two  polygons. 

418.  Def.   If  the   sides  of  one  polygon,  taken  in  order  as 
antecedents,  form  a  series  of  equal  ratios  with  the   sides  of 
another  polygon,  taken  in  order  as  consequents,  the  polygons 
are  said  to  have  their  sides  proportional.     Thus,  in  the  accom- 
panying figure,  if  a  :  I  =  b  :  m  =  c  :  n  —  d  :  o  =  e  :  p,  the  two 
polygons  have  their  sides  proportional.     The  lines  forming  any 
ratio  are  called  homologous  lines  of  the  two  polygons,  and  the 
ratio  of  two  such  lines  is  called  the  ratio  of  similitude  of  the 
polygons. 


419.    Def.     Two  polygons  are  similar  if  they  are  mutually 
equiangular  and  if  their  sides  are  proportional. 


BOOK   III 


177 


PROPOSITION  XIV.     THEOREM 

420.   Two  triangles  which  are  mutually  equiangular 
are  similar. 


H 


A  K        C 

Given   A  ABO  and  DEF  with 


=  Zz>,    Z  £  =  Z  #,  and 


To  prove    A  ABC  ~  A  DEF. 


ARGUMENT 

1.  Place  A  DEF  on  A  AB  c  so  that  Z  D  shall 

coincide  with  Z^4,  DE  falling  on  AB 
and  DF  on  .4(7.  Represent  A  DEF  in 
its  new  position  by  A  AHK. 

2.  Z  £  =  Z  £  =  Z 

3.  .\HK\\BC. 


5. 


6.  By  placing  A  D^  on  A  ABC  so  that 
Z.E  shall  coincide  with  Z.B,  it  may 
be  shown  that 

AB      BC 


7. 


AB__BC!__  AC 
'  DE      EF      DF 
8.  .'.  A  ABC~  A  DEF. 


Q.E.D. 


REASONS 
1.    §54,14. 


2.  By  hyp. 

3.  §  184. 

4.  §  409. 


5.  §  309. 

6.  By  steps  sim- 

ilar to  1-5. 


7.  §  54,  1, 

8.  §  419. 


421.  Cor.  i.  if  two  triangles  have  two  angles  of  one 
equal  respectively  to  two  angles  of  the  other,  the  triangles 
are  similar. 


178 


PLANE   GEOMETRY 


422.  Cor.  II.     Two  right  triangles  are  similar  if  an 
acute  angle  of  one  is  equal  to  an  acute  angle  of  the  other. 

423.  Cor.  in.   If  a  line  is  drawn  parallel  to  any  side  of 
a  triangle,  this  line,  with  the  other  two  sides,  forms  *  a 
triangle  which  is  similar  to  the  given  triangle. 


Ex.  663.   Upon  a  given  line  as  base  construct  a  triangle  similar  to  a 
given  triangle. 

Ex.  664.    Draw  a  triangle  ABC.     Estimate  the  lengths  of  its  sides. 
Draw  a  second  triangle  DEF  similar  to  ABC  and 
having  DE  equal  to  two  thirds  of  AB.     Compute 
DF  and  EF. 


Ex.  665.  Any  two  altitudes  of  a  triangle  are  to 
each  other  inversely  as  the  sides  to  which  they  are 
drawn. 


Ex.  666.  At  a  cer- 
tain hour  of  the  day  a 
tree,  BC,  casts  a  shadow, 
CA.  At  the  same  time 
a  vertical  pole,  ED, 
casts  a  shadow,  DF. 
What  measurements  are 
necessary  to  determine 
the  height  of  the  tree  ? 

Ex.   667.     If  CA  is  found  to  be  64  feet ;  DF,  16  feet ;  ED,  10  feet  ; 
what  is  the  height  of  the  tree  ? 

Ex.  668.  To  find  the  distance  across 
a  river  from  A  to  B,  a  point  C  was  located 
so  that  BC  was  perpendicular  to  AB  at 
B.  CD  was  then  measured  off  100  feet 
in  length  and  perpendicular  to  BC  at  C. 
The  line  of  sight  from  D  to  A  intersected 
BC  at  E.  By  measurement  CE  was 
found  to  be  90  feet  and  EB  210  feet. 
What  was  the  distance  across  the  river  ? 


B 


BOOK  III  179 

Ex.  669.  Two  isosceles  triangles  are  similar  if  the  vertex  angle  of  one 
equals  the  vertex  angle  of  the  other,  or  if  a  base  angle  of  one  equals  a 
base  angle  of  the  other. 

424.  It   follows  from   the   definition  of   similar   polygons, 
§  419,  and  from  Prop.  XIV  that: 

(1)  Homologous  angles  of  similar  triangles  are  equal. 

(2)  Homologous  sides  of  similar  triangles  are  proportional. 

(3)  Homologous  sides  of  similar  triangles  are  the  sides  opposite 
equal  angles. 

425.  Note.  In  case,  therefore,  it  is  desired  to  prove  four  lines  pro- 
portional, try  to  find  a  pair  of  triangles  each  having  two  of  the  given 
lines  as  sides.      If,  then,  these  triangles  can  be  proved   similar,  their 
homologous  sides  will  be  proportional.     By  marking  with  colored  crayon 
the  lines  required  in  the  proportion,  the  triangles  can  readily  be  found. 
If  it  is  desired  to  prove  the  product  of  two  lines  equal  to  the  product  of 
two  other  lines,  prove  the  four  lines  proportional  by  the  method  just  sug- 
gested, then  put  the  product  of  the  extremes  equal  to  the  product  of  the 
means.* 

426.  Def.    The  length  of  a  secant  from  an  external  point  to 
a  circle  is  the  length  of  the  segment  included  between   the 
point  and  the  second  point  of  intersection  of  the  secant  and 
the  circumference. 


Ex.  670.  If  two  chords  intersect  within  a  circle,  es- 
tablish a  proportionality  among  the  segments  of  the  chords. 
Place  the  product  of  the  extremes  equal  to  the  product  of 
the  means,  and  state  your  result  as  a  theorem. 

Ex.  671.  If  two  secants  are  drawn  from  any  given  point  to  a  circle, 
what  are  the  segments  of  the  secants?  Does  the  theorem  of  Ex.  670 
still  hold  with  regard  to  them? 

Ex.  672.  Kotate  one  of  the  secants  of  Ex.  671  about 
the  point  of  intersection  of  the  two  until  the  rotating 
secant  becomes  a  tangent.  What  are  the  segments  of  the 
secant  which  has  become  a  tangent  ?  Does  the  theorem  of 
Ex.  670  still  hold  ?  Prove. 

*  By  the  product  of  two  lines  is  meant  the  product  of  their  measure-numbers.  This 
will  be  discussed  again  in  Book  IV. 


180 


PLANE   GEOMETRY 


PROPOSITION  XV.     THEOREM 

427.     Two  triangles  which  have  their  sides  proportional 
are  similar. 


H 


K 


Given     A  DEF  and  ABC  such  that     —  = 

AB 
To  prove     A  DEF  ~  A  ABC. 

ARGUMENT 

1.  On  DE  lay  off  DH  —  AB,  and  on  DF  lay 

off  DK=AC. 

2.  Draw  HK. 

DE  _DF 
AB  AC 
DE  ^  DF 
DH~  DK 

5.  .'.  HK  II  EF. 

6.  .'.  A  DEF  ~  A  DHK. 

It  remains  to  prove  A  DHK  =  A  ABC. 

7.  —  ^-— , .  V-^-  ,  ^ 


3. 


4  .   VE  _  DF 


8. 


BJ3*        '  pjf'    BC^' 

9.          .:HK  =  BC. 

10.  Kow  J9^=  AB  and  DK=  AC. 

11.  .-.  A  J)HK=A  ABC. 

12.  But  A  7)£F~  A  Dfl-iT. 

13.  .-.  A  DEF  ~  A  ^#6".  Q.E.D. 


^(7  J:C7 

REASONS 

1.  §54,14. 

2.  §  54,  15. 

3.  By  hyp. 

4.  §309. 

5.  §415. 

6.  §423. 

7.  §  424,  2. 

8.  By  hyp. 

9.  §391. 

10.  Arg.  1. 

11.  §116. 

12.  Arg.  6. 

13.  §309. 


C 


Ex.  673.     Tf  the  sides  of  two  triangles  are  0,  12,  15,  and  6,  8,  10, 
respectively,  are  the  triangles  similar  ?     Explain. 


BOOK  III 


181 


Ex.  674.  Construct  a  triangle  that  shall  have  a  given  perimeter  and 
shall  be  similar  to  a  given  triangle. 

Ex.  675.  Construct  a  trapezoid,  given  the  two  bases  and  the  two 
diagonals.  HINT.  How  do  the  diagonals  of  a  trapezoid  divide  each  other  ? 

PROPOSITION  XVI.     THEOREM 

428.  If  two  triangles  have  an  angle  of  one  equal  to 
an  angle  of  the  other,  and  the  including  sides  propor- 
tional, the  triangles  are  similar. 


A  K        C        D 

Given  A  ABC  and  DEF  with  /.  A  =  * 
To  prove  A  AB  C  ~  A  DJ£F. 


_        -,  ^B       ^C 
D  and  —  =  —  • 
DE      DF 


3. 

4. 
5. 
6. 

7. 


ARGUMENT 

Place  A  DEF  on  A  45(7  so  that  Z  />  shall 
coincide  with  Z  ^4,  D-E  falling  on  AB, 
and   D-F   on    .4(7.     Represent  A  DEF 
in  its  new  position  by  A  AHK. 
AB  _AC 
DE      DF 
AB  _AC 
'  AH~~  AK 
.'.  HK  ||  BC. 
.'.  A  ABC  ~  A  AHK. 

But  A  AHK  is  A  DEF  transferred   to  a 
different  position. 

.-.  A  ABO  ~  A  DEF.  Q.E.D. 


REASONS 
1.    §  54,  14. 


2.  By  hyp. 

3.  §309. 


§415. 
§423. 
Arg.  1. 


7.    §309. 


Ex.  676.  Two  triangles  are  similar  if  two  sides  and  the  median  drawn 
to  one  of  these  sides  in  one  triangle  are  proportional  to  two  sides  and  the, 
corresponding  median  in  the  other  triangle. 


182 


PLANE   GEOMETRY 


B 


B 


D       C 

FIG.  1. 


E 

FIG.  2. 


AD  C 

FIG.  3. 


Ex.  677.     In  triangles  AB C  and  DB (7,  Fig.  1,AB  =  AC  and  BD  =  BC. 
Prove  triangle  ABC  similar  to  triangle  DU<7. 

Ex.678.     InVlg.2,AB:AC  =  AE:AD.    Prove  triangle  AB  (7  simi- 
lar to  triangle  ^IDJE1. 

Ex.  679.     If  in  triangle  ABC,  Fig.  3,  (L4  =  BC,  and  if  Z)  is  a  point 
such  that  CA  :  AB  =  AB  :  AD,  prove  AB  =  BD. 

Ex.  680.     Construct  a  triangle  similar  to  a  given  triangle  and  having 
the  sum  of  two  sides  equal  to  a  given  line. 


PROPOSITION  XVII.     THEOREM 

429.   Two  triangles  that  have  their  sides  parallel  each 
to  each,  or  perpendicular  each  to  each,  are  similar. 


FIG.  1. 


Given     A  ABC  and  A'B'C',  with  AB,  BC,  and  CA  II  (Fig.  1) 
or  _L  (Fig.  2)  respectively  to  A'B',  B'C',  and  C'A', 
To  prove     A  ABC  ~  A  A'B'C'. 


ARGUMENT 

1.  AB,  BC,  and  CA  are  II  or  _L  respectively 

to  ^'7?',  B'C',  and  Cy'.4f. 

2.  .*.  A  A,  B,  and  C  are  equal  respectively  or 

are  sup.  respectively  to  A  A',  B',  and  c'. 


REASONS 

1.  By  hyp. 

2.  §§198,201. 


BOOK  III 


183 


ARGUMENT 

3.  Three  suppositions  may  be  made,  there- 

fore, as  follows : 

=  2  rt.  A,  Z  C  +  Z  c'  =  2  rt.  A. 

Z  c  +  Z  (7'  =  2  rt.  A 
(3)   Z  J.  =  Z  ^',  Z£  =  Z  £';  hence, 
also,  Z  cy  ==  Z  <7'. 

4.  According  to  (1)  and  (2)  the  sum  of  the 

A  of  the  two  A  is  more  than  four  rt.  A. 

5.  But  this  is  impossible. 

6.  .•.  (3)  is  the  only  supposition  admissible; 

i.e.   the    two  A   are   mutually   equi- 
angular. 


7.   .-.  A  ABC~  AA'B'C'. 


Q.E.D. 


REASONS 


3.    §161,  a. 


4.    §  54,  2. 


5.  §204. 

6.  161,  b. 


7.    §420. 


430.  Question.     Can  one  pair  of  angles  in  Prop.  XVII  be  supple- 
mentary and  the  other  two  pairs  equal  ? 

SUMMARY    OF    CONDITIONS    FOB    SIMILARITY    OF    TRIANGLES 

431.  I.   Two  triangles  are  similar  if  they  are  mutually  equi- 
angular. 

(a)  Two  triangles  are  similar  if  two  angles  of  one  are  equal 
respectively  to  two  angles  of  the  other. 

(6)  Two  right  triangles  are  similar  if  an  acute  angle  of  one 
is  equal  to  an  acute  angle  of  the  other. 

(c)  If  a  line  is  drawn  parallel  to  any  side  of  a  triangle, 
this  line,  with  the  other  two  sides,  forms  a  triangle  which  is 
similar  to  the  given  triangle. 

II.  Two  triangles  are  similar  if  their  sides  are  proportional. 

III.  Two  triangles  are  similar  if  they  have  an  angle  of  one 
equal  to  an  angle  of  the  other,  and -the  including  sides  pro- 
portional. 

IV.  Two  triangles  are  similar  if  their  sides  are  parallel  each 
to  each,  or  perpendicular  each  to  each. 


184 


PLANE   GEOMETRY 


Ex.  681.  Inscribe  a  triangle  in  a  circle  and  circumscribe  about  the 
circle  a  triangle  similar  to  the  inscribed  triangle. 

Ex.  682.  Circumscribe  a  triangle  about  a  circle  and  inscribe  in  the 
circle  a  similar  triangle. 

Ex.  683.  The  lines  joining  the  mid-points  of  the  sides  of  a  triangle 
form  a  second  triangle  similar  to  the  given  triangle. 

Ex.  684.  ABC  is  a  triangle  inscribed  in  a  circle.  A  line  is  drawn 
from  A  to  P,  any  point  of  BC,  and  a  chord  is  drawn  from  B  to  a  point  Q 
in  arc  BC  so  that  angle  ABQ  equals  angle  APC.  Prove  AB  x  AC  = 
AQ  x  AP. 

PROPOSITION  XVIII.     THEOREM 

432.  The  bisector  of  an  angle  of  a  triangle  divides  the 
opposite  side  into  segments  which  are  proportional  to  the 
other  two  sides. 


A  PC 

Given  A  ABC  with  BP  the  bisector  of  Z  ABC. 
To  prove    AP :  PC  =  AB  :  B C. 


ARGUMENT 

1.  Through  C  draw  CE  II  PB,  meeting  AB 

prolonged  at  E. 

2.  In  A  AEG,  AP  :  PC  —  AB  :  BE. 

3.  Now  Z  3  =  Z  1. 

4.  AndZ4  =  Z2. 

5.  ButZl=Z2. 

6.  .-.  Z3  =  Z4. 

7.  .'.  BC=  BE. 

8.  .'.  AP  :  PC  =  AB  :  BC.          Q.E.D. 


REASONS 

1. 

§179. 

2. 

§410. 

3. 

§190. 

4. 

§189. 

5. 

By  hyp. 

6. 

§  54,  1. 

7. 

§162. 

8. 

§  309. 

BOOK   III 


185 


"El*-.  685.    The  sides  of  a  triangle  are  8,  12,  and  15.     Find  the  seg- 
ments of  side  8  made  by  the  bisector  of  the  opposite  angle. 

Ex.  6860     In  the  triangle  of  Ex.  685,  find  the  segments  of  sides  12  and 
15  made  by  the  bisectors  of  the  angles  opposite. 


PROPOSITION  XIX.     THEOREM 

433.  The  bisector  of  an  exterior  angle  of  a  triangle  di- 
vides the  opposite  side  externally  into  segments  which  are 
proportional  to  the  other  two  sides. 


Given  A  ABC,  with  BP  the  bisector  of  exterior  Z  CBF. 
To  prove    AP  :  PC  —  AB  :  BC. 


ARGUMENT 

1.  Through  C  draw  CE  II  PB,  meeting  AB 

a**. 

2.  Then  in  A  ABP,  AP  :  PC  =  AB  :  BE. 

Now  Z3  =  Z1. 
And  Z4  =  Z2. 
But  Z1=Z2. 

.-.  Z3  =  Z4. 

.'.  BC  =  BE. 
/.  AP  :  PC  =  AB  :  BC. 

Q.E.D. 


REASONS 

1. 

§179. 

2. 

§409. 

3. 

§190. 

4. 

§189. 

5. 

By  hyp. 

6. 

§  54,  1. 

7. 

§162. 

8. 

§309. 

Ex.  687.  Compare  the  lettering  of  the  figures  for  Props.  XVIII  and 
XIX,  and  also  the  steps  in  the  argument.  Could  one  argument  serxe  for 
the  two  cases  ? 

Ex.  688.  The  sides  of  a  triangle  are  9,  12,  and  16.  Find  the  segments 
of  side  9  made  by  the  bisector  of  the  exterior  angle  at  the  opposite  vertex. 


186 


PLANE   GEOMETRY 


434.  Def.  A  line  is  divided  harmonically  if  it  is  divided 
internally  and  externally  into  segments  whose  ratios  are  numeri- 
cally equal ;  thus,  if  line  AC  is  divided  internally  at  P  and  ex- 
ternally at  Q  so  that  the  ratio  of  (  f  t 

AP  to  PC  is  numerically  equal  to    ^  PC  Q 

the  ratio  of  AQ  to  QC,  AC  is  said  to  be  divided  harmonically. 


Ex.  689.  The  bisectors  of  the  interior  and  exterior  angles  at  any 
vertex  of  a  triangle  divide  the  opposite  side  harmonically. 

Ex.  690.  Divide  a  given  straight  line  harmonically  in  the  ratio  of 
3  to  6  ;  in  the  ratio  of  a  to  &,  where  a  and  b  are  given  straight  lines. 

PROPOSITION  XX.     THEOREM 

435.  In  two  similar  triangles  any  two  homologous  alti- 
tudes have  the  same  ratio  as  any  two  homologous  sides. 

A 


Given   two  similar  A  ABC  and  DEF,  with  two  corresponding 
altitudes  AH  and  DK. 


_  AH      AB       BC      CA 

To  prove    —  =  —  =  —  =  -r- , 

DK      DE      EF      FD 


ARGUMENT 

1.  In  rt.  A  ABH  and  DEK,  Z.  B  =  Z.E. 

2.  .-.  AABH~  A  DEK. 

o        AH,  opposite  Z  B_AB,  opposite  /.BHA 
'  '  DK,  opposite  Z.E    DE,  opposite  s/EKD 


^    ,  AB       BC       CA 
4.    But  —  =  —  =  —  . 
EF      FD 


DE 
AH  _  AB 
DK  ~  DE 


BC 
EF 


CA 


Q.E.D. 


REASONS 

1.  §424,1. 

2.  §422: 

3.  §  424,  2. 

4.  §424,2. 

5.  §  54,  1. 


BOOK  HI 


187 


PROPOSITION   XXI.     THEOREM 

436.  If  three  or  more  straight  lines  drawn  through  a 
common  point  intersect  two  parallels,  the  corresponding 
segments  of  the  parallels  are  proportional. 


/        I       \        \          /      I    \      \ 

Given  lines  PA,  PB,  PC,  PD  drawn  through  a  common  point  P 
and  intersecting  the  II  lines  AD  and  A'D'  at  points  Aj  B,  (7,  D  and 
A'j  B',  C1,  D',  respectively. 

SOL 

C'D'' 


A  7? 

To  prove   — - 
A'B' 


BC 
B'C' 

ARGUMENT 

1.  AD  II  A'D'. 

2.  .'.A  APE  ~  A  A'PB'. 

AB  _PB 
A'B'~  PB1' 


3. 


4.  Likewise  A  BPC 

5.  And    -*£=££ 


A  J?'PC'. 


B'C' 
AB 


PB' 
BC 


A'B'      B'C'' 

1.   Likewise  it  can  be  proved  that 
BC  _  CD 
B'C'~~  C'D'' 


8. 


AB 

'ATB' 


BC 
B'C' 


CD 

C'D' 


Q.E.D. 


REASONS 

1.  By  hyp. 

2.  §  423. 

3.  §  424,  2. 

4.  Args.  1-2. 

5.  §  424,  2. 

6.  §  54,  1. 

7.  By  steps  sim- 

ilar to  1-6. 

8.  §  54, 1. 


Ex.  691.    If  three  or  more  non-parallel  straight  lines  intercept  pro- 
portional segments  on  two  parallels,  they  pass  through  a  common  point. 


188 


PLANE   GEOMETRY 


Ex.  692.  A  man  is  riding  in  an  automobile  at  the  uniform  rate  of  30 
miles  an  hour  on  one  side  of  a  road,  while  on  a  footpath  on  the  other  side 
a  man  is  walking  in  the  opposite  direction.  If  the  distance  between  the 
footpath  and  the  auto  track  is  44  feet,  and  a  tree  4  feet  from  the  footpath 
continually  hides  the  chauffeur  from  the  pedestrian,  does  the  pedestrian 
walk  at  a  uniform  rate  ?  If  so,  at  what  rate  does  he  walk  ? 

Ex.  693.  Two  sides  of  a  triangle  are  8  and  11,  and  the  altitude  upon 
the  third  side  is  6.  A  similar  triangle  has  the  side  homologous  to  8  equal 
to  12.  Compute  as  many  parts  of  the  second  triangle  as  you  can. 

Ex.  694.  In  two  similar  triangles,  any  two  homologous  bisectors  are 
in  the  same  ratio  as  any  two  homologous  sides. 

Ex.  695.  In  two  similar  triangles,  any  two  homologous  medians  are 
in  the  same  ratio  as  any  two  homologous  sides. 


437.  Drawing  to  Scale.  Measure  the  top  of  your  desk. 
Make  a  drawing  on  paper  in  which  each  line  is  y1^  as  long  as 
the  corresponding  line  of  your  desk.  Check  your  work  by 
measuring  the  diagonal  of  your  drawing,  and  the  corresponding 
line  of  your  desk.  This  is  called  drawing  to  scale.  Map  draw- 
ing is  a  common  illustration  of  this  principle.  The  scale  of 
the  drawing  may  be  represented  :  (1)  by  saying,  "  Scale,  yL  "  or 
"  Scale,  1  inch  to  12  inches  "  ;  (2)  by  actually  drawing  the  scale 
as  indicated. 


Ex.  696.    Using  the  scale  above,  draw  lines  on  paper  to  represent 


24  inches  ;   3  feet  3  inches. 

Ex.  697.  On  the  black- 
board draw,  to  the  scale 
above,  a  circle  whose  diame- 
ter is  28  feet. 

Ex.  698.  The  figure 
represents  a  farm  drawn  to 
the  scale  indicated.  Find 
the  cost  of  putting  a  fence 
around  the  farm,  if  the  fenc- 
ing costs  $2.50  per  rod. 


B 


BOOK  III  189 

PROPOSITION  XXII.     THEOREM 

438.  If  two  polygons  are  composed  of  the  same  number 
of  triangles,  similar  each  to  each  and  similarly  placed, 
the  polygons  are  similar. 


H 


Given  polygons  ABODE  and  FGHIJ  with  A  ABC  ~  A  FGHj 
AACD  ~  A  FHI,  A  ADE  ~  A  FIJ. 

To  prove    polygon  ABODE  ~  polygon  FGHIJ. 

ARGUMENT  REASONS 

1.  In  A  ABC  and  FGH,  Z  B  =  Z  0.  1.    §  424,  1. 

2.  AlsoZl  =  Z2.  2.    §424,1. 

3.  In  A  ACD  and  OT7,  Z  3  =  Z4. 

4.  .-.  Z1  +  Z3  =  Z2  +  Z4. 

5.  .'.  Z  5CZ*  =  Z  (?#/. 

6.  Likewise  Z  (me  =  Z  HIJ,  Z  ^  =  Z  J, 

7.  .*.  polygons   ABODE    and    FGHIJ    are 

mutually  equiangular. 

8.  In  2! 


9. 


HF      HI 


IF 


10.   And  in  A  ADE  and  F/e7,  —  =  —  =  —. 

IF      IJ      JF 


EA 


11       •    AB  —  BC  —  CD  =  DE  — 

FG  ~~  GH  ~~  HI~  IJ  ~  JF' 

12.    .-.  polygon  ABODE  ~  polygon  FGHIJ. 

Q.E.D. 


3.  §  424,  1. 

4.  §  54,  2. 

5.  §  309. 

6.  By  steps  sim- 

ilar to  1-5. 

7.  By  proof. 


8.  §  424,  2. 

9.  §424,2. 

10.  §  424,  2. 

11.  §54,1. 

12.  §419. 


190 


PLANE   GEOMETRY 


439.  Cor.   Any  two  similar  polygons  may  be  divided 
into  the  same  number  of  triangles  similar  each  to  each 
and  similarly  placed. 

PROPOSITION  XXIII.     PROBLEM 

440.  Upon  a  line  homologous  to  a  side  of  a  given  poly- 
gon, to  construct  a  polygon  similar  to  the  given  polygon. 

C 

"--*— ^    D  /O 

P.-L 


Given   polygon  AD  and  line  JI/Q  homol.  to  side  AE. 
To  construct,    on  MQ,  a  polygon  ~  polygon  AD. 

I.    Construction 

1.  Draw  all  possible  diagonals  from  A,  as  AC  and  AD. 

2.  At  M,  beginning  with  MQ  as  a  side,  construct  A  7,  8,  and 
9  equal  respectively  to  A  1,  2,  and  3.     §  125. 

3.  At  Q,  with  MQ  as  a  side,  construct  Z 10  equal  to  Z  4,  and 
prolong  side  QP  until  it  meets  ML  at  P.     §  125. 

4.  At  P,  with  PM  as  a  side,  construct  Zll  equal  to  Z5,  and 
prolong  side  PO  until  it  meets  MR  at  0.     §  125. 

5.  At  O,  with  OM  as  a  side,  construct  Z12  equal  to  Z6,  and 
prolong  side  ON  until  it  meets  MF  at  N.     §  125. 

6.  MP  is  the  polygon  required. 


II.    Proof 


ARGUMENT 


1.  A  ADE  ~  A  MPQ,  A  ACD  ~  A  MOP,  and 
A  ABC  ~  A  MNO. 

2.  .-.  polygon  MP  ~  polygon  AD.         Q.E.D. 


REASONS 


1.    §  421. 


2.    §  438. 


BOOK  III  191 

III.    The  discussion  is  left  as  an  exercise  for  the  student. 

PROPOSITION  XXIV.     THEOREM 

441.   The  perimeters  of  two  similar  polygons  are  to  each 
other  as  any  two  homologous  sides. 

d 

e 


Given    ~  polygons  P  and  Q,  with  sides  a,  b,  c,  d,  e,  and  / 
hoinol.  respectively  to  sides  k,  I,  in,  n,  o,  and  p. 
perimeter  of  P  _a 


To  prove 


perimeter  of  Q      k 
ARGUMENT 


a 


k      I      m     n     o     p 
o     .   a  + &+ c -f-d+e+/_a 
fc-f-  I  -f-  m  H-  n  -f-  o  -\-p     k 

o     mi    4.  •     perimeter  of  P      a 
3.    I  hat  is,  i—  —  =  -. 

perimeter  oi  Q      k  Q.E.D. 


REASONS 

1.  §419. 

2.  §  401. 

3.  §309. 


Ex.  699.  The  perimeters  of  two  similar  polygons  are  152  and  138 ; 
a  side  of  the  first  is  8.  Find  the  homologous  side  of  the  second. 

Ex.  700.  The  perimeters  of  two  similar  polygons  are  to  each  other  as 
any  two  homologous  diagonals. 

Ex.  701.  The  perimeters  of  two  similar  triangles  are  to  each  other 
as  any  two  homologous  medians. 

Ex.  702.  If  perpendiculars  are  drawn  to  the  hypotenuse  of  a  right 
triangle  at  its  extremities,  and  if  the  other  two  sides  of  the  triangle  are 
prolonged  to  meet  these  perpendiculars,  the  figure  thus  formed  contains 
five  triangles  each  of  which  is  similar  to  any  one  of  the  others. 


192 


PLANE   GEOMETRY 


PROPOSITION  XXV.     THEOREM 

442.  In  a  right  triangle,   if  the  altitude  upon  the 
hypotenuse  is  drawn: 

I.  The  triangles  thus  formed  are  similar  to  the  given 
triangle  and  to  each  other. 

II.  The  altitude  is  a  mean  proportional  between  the 
segments  of  the  hypotenuse. 

III.  Either  side  is  a  mean  proportional  between  the 
whole  hypotenuse  and  the  segment  of  the  hypotenuse 
adjacent  to  that  side. 


f\  D 

Given   rt.  A  ABC  and  the  altitude  CD  upon  the  hypotenuse. 
To  prove : 

I.    A  ^CR^^ABC,  A  ADC  ~  A  ABC,  and  A  BCD  ~  A  ADC. 
II.    BD  :  CD  =  CD  :  DA. 
III.    AB  :  B C  =  BC :  DB  and  AB  :  AC  =  AC :  AD. 


I.  ARGUMENT 

1.  In  rt.  A  B  CD  and  ABC,  Z.B  =  Z.B. 

2.  .\A  BCD  ~  A  ABC.  N 

3.  In  rt.  A  ^Z>(7  and  ^B<7,  Z.A—/.A. 

4.  .-.  A  ^Z><?  ~  A  ABC. 

6.    .-.A  £C7D  ~  A  ADC.  Q.E.D. 


REASONS 

1.  By  iden. 

2.  §  422. 

3.  By  iden. 

4.  §  422. 

5.  §  424,  1. 

6.  §  422. 


II.   The  proof  of  II  is  left  as  an  exercise  for  the  student. 

HINT.  Mark  (with  colored  chalk,  if  convenient)  the  lines  required  in 
the  proportion.  Decide  which  triangles  will  furnish  these  lines,  and  use 
the  fact  that  homologous  sides  of  similar  triangles  are  proportional. 


BOOK  III  193 

III.   The  proof  of  III  is  left  as  an  exercise  for  the  student. 
HINT.     Use  the  same  method  as  for  II. 

443.  Cor.  I.    In  a  right  triangle,  if  the  altitude  upon 
the  hypotenuse  is  drawn: 

I.  The  square  of  the  altitude  is  equal  to  tfo  product  of 
the  segments  of  the  hypotenuse.    Thus,  CD2  =  BD  •  DA.     (See 
§  442,  II.) 

II.  The  square  of  either  side  is  equal  to  the  product 
of  the  whole  hypotenuse  and  the  segment  of  the  hypote- 
nuse adjacent  to  that  side.    Thus,  £C2=  AS  •  DB,  and  AC^= 
AB  -  AD.     (See  §  442,  III.) 

444.  Cor.  n.     If  from,  any  point  in  the  circumference 
of  a  circle  a  perpendicular  to  a 

diameter  is  drawn,  and  if  chords 
are  drawn  from  the  point  to  the 
ends  of  the  diameter  : 

I.  The  perpendicular  is  a  mean 
proportional  between  the  segments 
of  the  diameter. 

II.  Either  chord  is   a  mean  proportional  between  the 
whole  diameter  and  the  segment  of  the  diameter  adjacent 
to  the  chord. 

HINT.     A  APE  is  a  rt.  A.     Apply  Prop.  XXV,  II  and  III. 


Ex.  703.     In  a  right  triangle,  the  squares  of  the  two  sides  are  pro- 
portional to  the  segments  of  the  hypotenuse  made  by  the  altitude  upon  it. 
HINT.     Apply  §  443,  II. 

Ex.   704.     The  sides  of  a  right  triangle  are  9,  12,  and  15. 

(1)  Compute  the  segments  of  the  hypotenuse  made   by  the  altitude 
upon  it. 

(2)  Compute  the  length  of  the  altitude. 

Ex.  705.  If  the  two  arms  of  a  right  triangle  are  6  and  8,  compute  the 
length  of  the  perpendicular  from  the  vertex  of  the  right  angle  to  the 
hypotenuse. 


194  PLANE   GEOMETRY 


PROPOSITION  XXVI.     PROBLEM 

445.    To  construct   a  mean  proportional  between  two 
given  lines. 


/ 

/        fiol    i     f 

A rn ? C « 0# 

Given  two  lines,  m  and  w. 

To  construct   a  line  I  so  that  m:l  =  I:  n. 

I.    Construction 

1.  On  any  indefinite  str.  line,  as  -4-B,  lay  off  AC  =  m  and 
CD  =  w. 

2.  With  0,  the  mid-point  of  ^Z>,  as  center  and  with  a  radius 
equal  to  07),  describe  a  semicircumference. 

3.  At  C  construct  CEA.AD,  meeting  the  semicircumference 
at  E.     §  148. 

4.  CE  is  the  required  line  I. 

II.  The  proof  and  discussion  are  left  as  an  exercise  for  the 
student.  

Ex.  706.  By  means  of  §  444,  II,  construct  a  mean  proportional  be- 
tween two  given  lines  by  a  method  different  from  that  given  in  Prop. 
XXVI. 

Ex.  707.  Use  the  method  of  Prop.  XXVI  to  construct  a  line  equal 
to  \/3  «6,  a  and  b  being  given  lines. 

ANALYSIS.        Let  x  =  the  required  line. 

Then  x  =  V3  ab. 

.-.  x2  =  3  ab. 
.'.  3  a  :  x  =  x  :  b. 

Ex.  708.     Construct  a  line  equal  to  a\/3,  where  a  is  a  given  line. 

Ex.  709.  Using  a  line  one  inch  long  as  a  unit,  construct  a  line  equal 
to  V3 ;  \/5  ;  V(>.  Choosing  your  own  unit,  construct  a  line  equal  to 
3v%  2V3,  5V6. 


BOOK  in 


195 


PROPOSITION   XXVII.     THEOREM 

446.  In  any  right  triangle  the  square  of  the  hypotenuse 
is  equal  to  the  sum  of  the  squares  of  the  other  two  sides. 

C 


Given     rt.  A  ABC,  with  its  rt.  Z  at  (7. 
To  prove     a2  -f-  b2  =  c2. 

ARGUMENT 

1.    From    C  draw    CD  J_  AB   forming   seg 
ments  I  and  d. 


a2  =  c  •  L 


+  b2  =  c  •  c  =  c2. 


Q.E.D. 


REASONS 

1.  §155. 

2.  §  443,  II. 

3.  §443,11. 

4.  §54,2. 

5.  §309. 


447.  Cor.  I.     The  square  of  either  side  of  a  right  tri- 
angle is  equal  to  the  square  of  the  hypotenuse  minus  the 
square  of  the  otJier  side. 

448.  Cor.  II.    The  diagonal  of  a  square 
is  equal   to  its  side1,  multiplied  by  tlie 
square  root  of  two. 


OUTLINE   OF   PROOF 


2. 


Q.E.D. 


449.  Historical  Note.  The  property  of  the  right  triangle  stated 
in  Prop.  XXVII  was  known  at  a  very  early  date,  the  ancient  Egyptians, 
2000  B.C.,  having  made  a  right  triangle  by  stretching  around  three  pegs  a 
cord  measured  off  into  3,  4,  and  5  units.  See  Note,  Book  IV,  §  510. 


196  PLANE   GEOMETRY 

Ex.  710.     By  means  of  §  448,  construct  a  line  equal  to  V2  inches. 

Ex.   711.     If  a  side  of  a  square  is  6  inches,  find  its  diagonal. 

Ex.  712.  The  hypotenuse  of  a  right  triangle  is  15  and  one  arm  is  9. 
Find  the  other  arm  and  the  segments  of  the  hypotenuse  made  by  the 
perpendicular  from  the  vertex  of  the  right  angle. 

Ex.  713.  Find  the  altitude  of  an  equilateral  triangle  whose  side  is  6 
inches. 

Ex.  714.  Find  a  side  of  an  equilateral  triangle  whose  altitude  is 
8  inches. 

Ex.  715.  Divide  a  line  into  segments  which  shall  be  in  the  ratio 
of  1  to  V2. 

Ex.  716.  The  radius  of  a  circle  is  10  inches.  Find  the  length  of  a 
chord  6  inches  from  the  center  ;  4  inches  from  the  center. 

Ex.  717.  The  radius  of  a  circle  is  20  inches.  How  far  from  the  cen- 
ter is  a  chord  whose  length  is  32  inches  ?  whose  length  is  28  inches  ? 

Ex.  718.  In  a  circle  a  chord  24  inches  long  is  5  inches  from  the  center. 
How  far  from  the  center  is  a  chord  whose  length  is  12  inches  ? 


450.  Def.     The  projection  of  a  point  upon  a  line  is  the  foot 
of  the  perpendicular  from  the  point  to  the  line. 

451.  Def.     The  projection  of  a  line  segment  upon  a  line  is 
the  segment  of  the  second  line  included  between  the  projections 
of  the  extremities  of  the  first  line  upon  the  second. 

Thus,  C  is  the  projection  of  A  upon  MN,  D  is  the  projection 
of  B  upon  MN,  and  CD  is  the  projection  of  AB  upon  MN. 

K  B 


c     /      \D 


MC  D  N  M    C  D  N  M 


Ex.  719.  In  the  figures  above,  under  what  condition  will  the  projec- 
tion of  AB  on  MN  be  a  maximum  ?  a  minimum  ?  Will  the  projection 
CD  ever  be  equal  to  AB  ?  greater  than  AB  ?  Will  the  projection  ever 
be  a  point  ? 

Ex.  720.  In  a  right  isosceles  triangle  the  hypotenuse  of  which  is  10 
inches,  find  the  length  of  the  projection  of  either  arm  upon  the  hypotenuse. 

Ex.  721.  Find  the  projection  of  one  side  of  an  equilateral  triangle 
upon  another  if  each  side  is  6  inches. 


BOOK   ITT 


197 


Ex.  722.  Draw  the  projections  of  the  shortest  side  of  a  triangle  upon 
each  of  th3  other  sides  :  (1)  in  an  acute  triangle  ;  (2)  in  a  right  triangle  ; 
(3)  in  an  obtuse  triangle.  Draw  the  projections  of  the  longest  side  in 
each  case. 

Ex.  723.  Two  sides  of  a  triangle  are  8  and  12  inches  and  their 
included  angle  is  60°.  Find  the  projection  of  the  shorter  upon  the  longer. 

Ex.  724.  In  Ex.  723,  find  the  projection  of  the  shorter  side  upon  the 
longer  if  the  included  angle  is  30°  ;  45°. 

Ex.  725.  Parallel  lines  that  have  equal  projections  on  the  same  line 
are  equal. 

PROPOSITION  XXVIII.     PROBLEM 

452.  In  any  triangle  to  find  the  value  of  the  square  of 
the  side  opposite  an  acute  angle  in  terms  of  the  other  two 
sides  and  of  the  projection  of  either  of  these  sides  upon  the 


other. 


X     D 


FIG.  1. 


P 

FIG.  2. 


X 


Given  ABAX,  with  Z.X  acute;    and  p,  the  projection  of  b 
upon  a.  • 

To  find   the  value  of  y?  in  terms  of  a,  6,  and  p. 


ARGUMENT 

1.  In  rt.  A  BAD,  x?  —  AD2  +  D~B2. 

2.  But  AD2  =  b2  -  p2. 

3.  And  DB  =  a—  p  (Fig.  1)  or  p  —  a  (Fig.  2). 


=  a2  -  2  ap  +  p2. 


REASONS 

1.  §  446. 

2.  §  447. 

3.  §54,11. 

4.  §  54,  13. 

5.  §309. 


i.e.  x2  =  a2  -f  b2  —  2  ap.  Q.E.F. 

453.    Question.      Why  is  it  not  necessary  to  include  here  the  figure 
and  discussion  for  a  right  triangle  ? 


198 


PLANE   GEOMETRY 


454.  Prop.  XXVIII  may  be  stated  in  the  form  of  a 
theorem  as  follows : 

In  any  triangle,  the  square  of  the  side  opposite  an  acute  angle 
is  equal  to  the  sum  of  the  squares  of  the  other  two  sides  dimin- 
ished by  twice  the  product  of  one  of  these  sides  and  the  projec- 
tion of  the  other  side  upon  it. 


Ex.  726.    If  the  sides  of  a  triangle  are  7,  8,  and  10,  is  the  angle 
opposite  10  obtuse,  right,  or  acute  ?    Why  ? 

Ex.  727.     Apply  the  statement  of  Prop.  XXVIII  to  the  square  of  an 
arm  of  a  right  triangle. 

Ex.  728.     Find  x  (in  the  figure  for  Prop.  XXVIII)  in  terms  of  a  and 
b  and  the  projection  of  a  upon  b. 

Ex.  729.     If  the  sides  of  a  triangle  are  13,  14,  and  15,  find  the  pro- 
jection of  the  first  side  upon  the  second. 

Ex.  730.     If  two  sides  of  a  triangle  are  4  and  12  and  the  projection  of 
the  first  side  upon  the  second  is  2,  find  the  third  side  of  the  triangle. 


PROPOSITION   XXIX.     THEOREM 

455.  In  any  obtuse  triangle,  the  square  of  the  side  oppo- 
site the  obtuse  angle  is  equal  to  the  sum  of  the  squares  of 
the  other  two  sides  increased  by  twice  the  product  of  one 
of  these  sides  and  the  projection  of  the  other  side  upon  it. 


Given   A  BAX  with  /.X  obtuse,  and  p,  the  projection  of  b 
upon  a. 

To  prove   x*  =  a2  +  b2  +  2  ap. 


BOOK  III  199 


ARGUMENT 

1.  In  rt.  A  BAD,  of  =  AD2  -f-  I)ff. 

2.  But    AD2  =  b2-p2. 

3.  And    DB  =  a+p. 

5.  .-.  »2  =  62-p2+a2  + 

i.e.     a^  =  a2  +  62  +  2  op.  Q.E.D. 


REASONS 


1.  §446. 

2.  §447. 

3.  §  54,  11. 

4.  §54,13. 

5.  §309. 


456.  From  Props.  XXVIII  and  XXIX,  we  may  derive  the 
following  formulas  for  computing  the  projection  of  one  side 
of  a  triangle  upon  another ;  thus  if  a,  6,  and  c  represent  the 
sides  of  a  triangle  : 

From  Prop.  XXVIII,    p  =  a'  +  W~c\  (1) 

From  Prop.  XXIX,   -  p  =  a*  +  b*  ~  c\  (2) 

£  a 

It  is  seen  that  the  second  members  of  these  two  equations 
are  identical  and  that  the  first  members  differ  only  in  sign. 
Hence,  formula  (1)  may  always  be  used  for  computing  the 
length  of  a  projection.  It  need  only  be  remembered  that  if  p 
is  positive  in  any  calculation,  it  indicates  that  the  angle  oppo- 
site c  is  acute;  while  if  p  is  negative,  the  angle  opposite  c 
is  obtuse.  It  can  likewise  be  shown,  (see  Prop.  XXVII)  that 
if  j>  =  0,  the  angle  opposite  c  is  a  right  angle. 


Ex.  731.     Write  the  formula  for  the  projection  of  a  upon  6. 

Ex.  732.  In  triangle  ABC,  a  =  15,  b  =  20,  c  =  25 ;  find  the  projec- 
tion of  6  upon  c.  Is  angle  A  acute,  right,  or  obtuse  ? 

Ex.  733.  In  the  triangle  of  Ex.  732,  find  the  projection  of  a  upon  b. 
Is  angle  C  acute,  right,  or  obtuse  ? 

Ex.  734.  The  sides  of  a  triangle  are  8,  14,  and  20.  Is  the  angle 
opposite  the  side  20  acute,  right,  or  obtuse  ? 

Ex.  735.  If  two  sides  of  a  triangle  are  10  and  12,  and  their  included 
angle  is  120°,  what  is  the  value  of  the  third  side  ? 

Ex.  736.*  If  two  sides  of  a  triangle  are  12  and  16,  and  their  included 
angle  is  45°,  find  the  third  side. 


200 


PLANE   GEOMETRY 


Ex.  737.     If  in  triangle  ABC,  angle  C  =  120°,  prove  that 
AB2  =  BC2  +  AC*  +  AC-BC. 

Ex.  738.  If  a  line  is  drawn  from  the  vertex  C  of  an  isosceles  tri- 
angle ABC,  meeting  base  AB  prolonged  at  D,  prove  that 

~CD2  _  CB2  =  AD  •  BD. 

PROPOSITION   XXX.     THEOREM 

457.  In  any  triangle,  the  sum  of  the  squares  of  any  two 
sides  is  equal  to  twice  the  square  of  half  the  third  side 
increased  by  twice  the  square  of  the  median  upon  that 
side. 


D 


B 


Given   A  ABC  with  ma,  the  median  to  side  a. 
To  prove   b2  +  c2  =  2f  ^Y+  2  ma2. 


AKGUMENT 

1.  Suppose   b  >  c ;  then  Z  ADC  is  obtuse 

and  Z  BDA  is  acute. 

2.  Let  p  be  the  projection  of  ma  upon  a. 

3.  Then  from  AJZ)C', 


4.   And  from 

C2  = 


REASONS 

1.  §  173. 

2.  §451. 

3.  §455. 


4.    §454. 


Q.E.D.      5.    §  54,  2. 


BOOK  III  201 

458.  Cor.     The  difference  of  the  squares  of  two  sides 
of  any  triangle  is  equal  to  twice  the  product  of  the  third 
side  and  the  projection  of  the  median  upon  the  third  side. 

HINT.     Subtract  the  equation  in  Arg.  4  from  that  in  Arg.  3  (§  457) 
member  from  member. 

Ex.  739.     Write  the  formula  involving  the  median  to  6  ;  to  c. 
Ex.  740.     Apply  Prop.  XXX  to  a  triangle  right-angled  at  A  ;  at  B  ; 
at.  C. 

459.  It  will  be  seen  that  the  formula  of  Prop.  XXX  con- 
tains the  three  sides  of  a  triangle  and  a  median  to  one  of  these 
sides.     Hence,  if  the  three  sides  of  a  triangle  are  given,  the 
median  to  any  one  of  them  can  be  found ;  also,  if  two  sides 
and  any  median  are  given,  the  third  side  can  be  found. 


Ex.  741.  If  the  sides  of  a  triangle  ABC  are  5,  7,  and  8,  find  the 
lengths  of  the  three  medians. 

Ex.  742.  If  the  sides  of  a  triangle  are  12,  16,  and  20,  find  the  median 
to  side  20.  How  does  it  compare  in  length  with  the  side  to  which  it  is 
drawn  ?  Why  ? 

Ex.  743.     In  triangle  ABC,  a  -  16,  b  -  22,  and  mc  =  17.     Find  c. 

Ex.  744.  In  a  right  triangle,  right-angled  at  C,  mc  =  8£ ;  what  is  c  ? 
Find  one  pair  of  values  for  a  and  b  that  will  satisfy  the  conditions  of  the 
problem. 

Ex.  745.  The  sum  of  the  squares  of  the  four  sides  of  any  parallel- 
ogram is  equal  to  the  sum  of  the  squares  of  its  diagonals. 

Ex.  746.  The  sum  of  the  squares  of  the  four  sides  of  any  quadrilateral 
is  equal  to  the  sum  of  the  squares  of  its  diagonals  increased  by  four  times 
the  square  of  the  line  joining  their  mid-points. 

Ex.   747.     Construct  a  triangle  ABC,  given  &,  c,  and  ma. 

Ex.  748.  Construct  a  triangle  ABC,  given  a,  6,  and  the  projection 
of  b  upon  a. 

Ex.  749.  Compute  the  side  of  a  rhombus  whose  diagonals  are  12 
and  10. 

Ex.  750.  If  the  square  of  the  longest  side  of  a  triangle  is  greater 
than  the  sum  of  the  squares  of  the  other  two  sides,  is  the  triangle  obtuse, 
right,  or  acute  ?  Why  ? 


202  PLANE   GEOMETRY 

PROPOSITION  XXXI.     THEOREM 

460.  //  through  a  point  within  a  circle  two  chords  are 
drawn,  the  product  of  the  two  segments  of  one  of  these 
chords  is  equal  to  the  product  of  the  two  segments  of  the 
other. 


Given   P,  a  point  within  circle  0,  arid  AB  and  CD,  any  two 
chords  drawn  through  P. 
To  prove    PA  •  PB  —  PC  -  PD. 
The  proof  is  left  as  an  exercise  for  the  student. 
HINT.    Prove  A  APG  ~  A  PDB. 


Ex.  751.  In  the  figure  for  Prop.  XXXI,  if  PA  =  5,  PB  =  12,  and 
PD  =  6,  find  PC. 

Ex.  752.  In  the  same  figure,  if  PC  =  10,  PD  =  8,  and  AB  =  21,  find 
PA  and  PS. 

Ex.  753.  In  the  same  figure,  if  PC  =  6,  DC  =  22,  and  AB  =  20, 
find  AP  and  PS. 

Ex.  754.  In  the  same  figure,  if  PA  =  m,  PC  =  n,  and  PD  =  r,  find 
PS. 

Ex.  755.  If  two  chords  intersecting  within  a  circle  are  of  lengths  8 
and  10,  and  the  second  bisects  the  first,  what  are  the  segments  of  the 
second  ? 

Ex.  756.  By  means  of  Prop.  XXXI  construct  a  mean  proportional 
between  two  given  lines. 

Ex.  757.  If  two  chords  intersect  within  a  circle  and  the  segments  of 
one  chord  are  a  and  b  inches,  while  the  second  chord  measures  d  inches, 
construct  the  segments  of  the  second  chord. 

HINT.     Find  the  locus  of  the  mid-points  of  chords  equal  to  d. 


BOOK   III 


203 


Ex.  758.  If  two  lines  AS  and  CD  intersect  at  E  so  that  AE  -  EB 
=  CE  •  ED,  then  a  circumference  can  be  passed  through  the  four  points 
A,  B,  C,  D. 

PROPOSITION  XXXII.     THEOREM 

461.  //  a  tangent  and  a  secant  are  drawn  from  any 
given  point  to  a  circle,  the  tangent  is  a  mean  propor- 
tional between  the  whole  secant  and  its  external  segment. 


Given  tangent  PT  and  secant  PB  drawn  from  the  point  P 
to  the  circle  0. 


m 

To  prove    — : 
PT 


PT 
PC 


1. 
2. 
3. 
4. 

5. 


ARGUMENT 
Draw  CT  and  BT. 
In  A  PET  and  CTP,  Z  P  =  Z  P. 
Z2  =  Z2'. 
.-.  APBT~  A  CTP. 

PB,  opposite  Z  3  in  A  PBT 

'  PT,  opposite  Z  3'  in  A  CTP 

_  PT,  opposite  Z  2  in  A  PBT 

PC,  opposite  Z  2'  in  A  CTP 


Q.E.D. 


REASONS 

1. 

§  54,  15. 

2 

By  iden. 

3. 

§  362,  a. 

4. 

§421.  ~ 

5. 

§  424,  2. 

462.  Cor.  I.  If  a  tangent  and  a  secant  are  drawn  from 
any  given  point  to  a  circle-,  the  square  of  the  tangent  is 
equal  to  the  product  of  tJie  whole  secant  and  its  external 
segment. 


204 


PLANE   GEOMETRY 


463.  Cor.  II.  If  two  or  more  secants  are  drawn  from 
any  given  point  to  a  circle,  the  product  of  any  secant  and 
its  external  segment  is  constant. 


Given  secants  PD,  PE,  PF, 
drawn  from  point  P  to  circle  0, 
and  let  their  external  segments 
be  denoted  by  PA,  PB,  P<7, 
respectively. 

To  prove 

PD  •  PA  =  PE  •  PB  =  PF  •  PC. 

ARGUMENT 

1.  From  P  draw  a  tangent  to  circle  0,  as 

PT. 

2.  PT2  =  PD  •  PA  ;    PT2  =  PE  •  PB ;    PT2  = 

PF  •  PC. 

3.  .'.  PD  •  PA  =  PE-  PB  —  PF  •  PC.       Q.E.D. 


D 


1.    §286. 


2.    §462. 


3.    §54,1. 


Ex.  759.  If  a  tangent  and  a  secant  drawn  from  the  same  point  to  a 
circle  measure  6  and  18  inches,  respectively,  how  long  is  the  external 
segment  of  the  secant  ? 

Ex.  760.  Two  secants  are  drawn  to  a  circle  from  an  outside  point.  If 
their  external  segments  are  12  and  9,  and  the  internal  segment  of  the  first 
secant  is  8,  what  is  the  length  of  the  second  secant  ? 

Ex.  761.  The  tangents  to  two  intersecting  circles  from  any  point  in 
their  common  chord  (prolonged)  are  equal. 

Ex.  762.  If  two  circumferences  intersect,  their  common  chord  (pro- 
longed) bisects  their  common  tangents. 


464.  Def.  A  line  is  said  to  be  divided  in  extreme  and  mean 
ratio  if  it  is  divided  into  two  parts  such  that  one  part  is  a 
mean  proportional  between  the  whole  line  and  the  other  part. 

Thus,  AB  is  divided  in  extreme  PR 

and  mean  ratio  at  P  if  AB  \  AP  =      « 

AP ;  PB.     This  division  is  known  as  the  golden  section. 


BOOK  III 


205 


PROPOSITION  XXXIII.     PIOBLEM 

465.   To  divide  a  line  internally  in  extreme  and  mean 
ratio. 


XX 

/ 


„-.*•' 


\  \\ 

i»    iV 


' 


A  P  B 

Given   line  AB. 

To  find,   in  AB,  a  point  P  such  that  AB  :  AP  =  AP  :  PB. 


I.    Construction 

1.  From  B  draw  SO  _L  .4B  and  =  \AB.     §  148. 

2.  With  0  as  center  and  with  BO  as  radius  describe  a  cir- 
cumference. 

3.  From  A  draw  a  secant  through  0,  cutting  the  circumfer- 
ence at  C  and  Z). 

4.  With  A  as  center  and  with  AC  as  radius  draw  CP,  cutting 
^45  at  P. 

5.  AB  :  AP  =  AP  :  P£. 


II.    Proof 
ARGUMENT 

1.  AB  is  tangent  to  circle  O. 

2.  .-.  AD  :  AB  =  AB  :  AC. 

3.  .'.  AD  —  AB  :  AB  =  AB  —  AC  :  AC. 

4.  .'.  AD  —  CD  :  AB  =  AB  —  ^P  :  AP. 

5.  .'.  AP:  ^/?  =  PJ3~:  AP. 

6.  .'.  ^£:  ^P  =  ^P:  PB. 


REASONS 

1.  §314. 

2.  §461. 

3.  §399_ 

4.  §  809. 

5.  §  309. 

6.  §  395. 


Q.E.D. 

III.    The  discussion  is  left  as  an  exercise  for  the  student. 


206 


PLANE   GEOMETRY 


Ex.  763.     Divide  a  line  AB  externally  in  extreme  and  mean  ratio. 
HINT.     In  the  figure  for  Prop.   XXXIII  prolong  BA  to  J",  making 
P'A  =  AD.    Then  prove  AB  -.  P'A  =  P'A  :  P'B. 

Ex.  764.  If  the  line  I  is  divided  internally  in  extreme  and  mean  ratio, 
and  if  s  is  the  greater  segment,  find  the  value  of  s  in  terms  of  I. 

HINT.     I :  s  =  s  :  I  —  s. 

Ex.  765.  A  line  10  inches  long  is  divided  internally  in  extreme  and 
mean  ratio.  Find  the  lengths  of  the  two  segments. 

Ex.  766.  A  line  8  inches  long  is  divided  externally  in  extreme  and 
mean  ratio.  Find  the  length  of  the  longer  segment. 


MISCELLANEOUS   EXERCISES 

Ex.  767.  Explain  how  the  accompanying 
figure  can  be  used  to  find  the  distance  from 
A  to  B  on  opposite  sides  of  a  hill.  CE  =  \BC, 
CD  =  \  AC.  ED  is  found  by  measurement  to 
be  125  feet.  What  is  the  distance  AB  ? 

Ex.  768.  A  little  boy  wished  to  obtain  the 
height  of  a  tree  in  his  yard.  He  set  up  a  ver- 
tical pole  6  feet  high  and  watched  until  the 
shadow  of  the  pole  measured  exactly  6  feet.  He 
then  measured  quickly  the  length  of  the  tree's  shadow  and  called  this  the 
height  of  the  tree.  Was  his  answer  correct  ?  Draw  figures  and  explain. 
Use  this  method  for  measuring  the  height  of  your  school  building  and  flag 
pole. 

Ex.  769.  If  light  from  a  tree, 
as  AB,  is  allowed  to  pass  through  a 
small  aperture  0,  in  a  window  shutter 
W,  and  strike  a  white  screen  or  wall, 
an  inverted  image  of  the  tree,  as  CD, 
is  formed  on  the  screen.  If  the  dis- 
tance OE=  30  feet,  OF  =  8  feet,  and 
the  length  of  the  tree  AB= 35  feet,  find 
the  length  of  the  image  CD.  Under 
what  condition  will  the  length  'of  the  image  equal  the  length  of  the  tree  ? 

This  exercise  illustrates  the  principle  of  the  photographer's  camera. 

Ex.  770.     By  means  of  Prop.  XXXII  construct  a  mean  proportional 
between  two  given  lines. 

Ex.  771.    In  a  certain  circle  a  chord  5\/5  inches  from  the  center  is  20 
inches  in  length.     Find  the  length  of  a  chord  9  inches  from  the  center. 


BOOK  III  207 

Ex.  772.  Compute  the  length  of  :  (1)  the  common  external  tangent, 
(2)  the  common  internal  tangent,  to  two  circles  whose  radii  are  8  and 
6,  respectively,  and  the  distance  between  whose  centers  is  20. 

Ex.  773.  If  the  hypotenuse  of  an  isosceles  right  triangle  is  16  inches, 
what  is  the  length  of  each  arm  ? 

Ex.  774.  If  from  a  point  a  tangent  and  a  secant  are  drawn  and  the 
segments  of  the  secant  are  4  and  12,  how  tong  is  the  tangent  ? 

Ex.  775.    Given  the  equation  m  +  n  =  —  ;  solve  for  x. 

X  +  C  C 

Ex.  776.  Find  a  mean  proportional  between  a2  +  2  ab  +  62  and 
a2  -  2  ab  +  62. 

Ex.  777.  The  mean  proportional  between  two  unequal  lines  is  less 
than  half  their  sum. 

Ex.  778.  The  diagonals  of  a  trapezoid  divide  each 
other  into  segments  which  are  proportional. 

Ex.  779.  ABC  is  an  isosceles  triangle  inscribed  in 
a  circle.  Chord  BD  is  drawn  from  the  vertex  B,  cutting 
the  base  in  any  point,  as  E.  Prove  BD  :  AB—AB  :  BE.  ""  —  ""/) 

Ex.  780.  In  a  triangle  ABO  the  side  AB  is  305  feet.  If  a  line 
parallel  to  B  C  divides  AC  in  the  ratio  of  2  to  3,  what  are  the  lengths  of  the 
segments  into  which  it  divides  AB  ? 

Ex.  781.  Construct,  in  one  figure,  four  lines  whose  lengths  shall  be 
that  of  a  given  unit  multiplied  by  V2,  \/3,  2,  Vs"",  respectively. 

Ex.  782.  Two  sides  of  a  triangle  are  12  and  18  inches,  and  the  perpen- 
dicular upon  the  first  from  the  opposite  vertex  is  9  inches.  What  is  the 
length  of  the  altitude  upon  the  second  side  ? 

Ex.  783.     If  a  :  b  =  c  :  d,  show  that 


. 

Also  translate  this  fact  into  a  verbal  statement. 

Ex.  784.  If  a  constant  is  added  to  or  subtracted  from  each  term  of 
a  proportion,  will  the  resulting  numbers  be  in  proportion  ?  Give  proof. 

Ex.  785.     If  r  :  s  =  t  :  g,  is  3  r  +  -  :  s  =  1  1  :  2  q  ?    Prove  your  answer. 

Ex.  786.  One  segment  of  a  chord  drawn  through  a  point  7  units  from 
the  center  of  a  circle  is  4  units  long.  If  the  diameter  of  the  circle  is  15 
units,  what  is  the  length  of  the  other  segment  ? 

Ex.  787.  The  non-parallel  sides  'of  a  trapezoid  and  the  line  joining 
the  mid-points  of  the  parallel  sides,  if  prolonged,  are  concurrent. 


208 


PLANE   GEOMETRY 


Ex.  788.  Construct  a  circle  which  shall  pass  through  two  given 
points  and  be  tangent  to  a  given  straight  line. 

Ex.  789.  The  sides  of  a  triangle  are  10,  12, 15.  Compute  the  lengths 
of  the  two  segments  into  which  the  least  side  is  divided  by  the  bisector 
of  the  opposite  angle.  . 

Ex.  790.  AB  is  a  chord  of  a  circle,  and  CE  is  any 
chord  drawn  through  the  middle  point  C  of  arc  AB,  cut- 
ting chord  AB  at  D.  Prove  GE :  CA  =  CA  :  CD. 

Ex.  791.  Construct  a  right  triangle,  given  its  perime- 
ter and  an  acute  angle. 

Ex.  792.  The  base  of  an  isosceles  triangle  is  a,  and  the  perpendicular 
let  fall  from  an  extremity  of  the  base  to  the  opposite  side  is  6.  Find  the 
lengths  of  the  equal  sides. 

Ex.  793.  AD  and  BE  are  two  altitudes  of  triangle  CAB.  Prove  that 
AD:  BE=  CA-.BC. 

Ex.  794.  If  two  circles  touch  each  other,  their  common  external 
tangent  is  a  mean  proportional  between  their  diameters. 

Ex.  795.  If  two  circles  are  tangent  internally,  all  chords  of  the  greater 
circle  drawn  from  the  point  of  contact  are  divided  proportionally  by  the 
circumference  of  the  smaller  circle. 

Ex.  796.  If  three  circles  intersect  each  other,  their  common  chords 
pass  through  a  common  point. 

Ex.  797.  The  square  of  the  bisector  of  an  angle  of  a  triangle  is  equal 
to  the  product  of  the  sides  of  this  angle  diminished  by  the  product  of  the 
segments  of  the  third  side  made  by  the  bisector. 

Given  A  ABC  with  t,  the  bisector  of  Z  J?,  dividing 
side  6  into  the  two  segments  s  and  r. 
To  prove  &  =  ac  —  rs. 

OUTLINE  OF  PROOF 


3. 


Prove  that  a  :  t  =  t-\-  m  :  c. 
Then  ac  =  i2  +  tm  =  t2  +  rs. 
.'.  t2  =  ac  —  rs. 


Ex.  798.  In  any  triangle  the  product  of  two 
sides  is  equal  to  the  product  of  the  altitude  upon  the 
third  side  and  the  diameter  of  the  circumscribed 
circle. 

HINT.     Prove  A  ABD  ^  l^EBC.     Then  prove 
ac  =  hd. 


BOOK  IV 


AREAS    OF    POLYGONS 

466.  A  surface  may  be  measured  by  finding  how  many  times 
it  contains  a  unit  of  surface.     The  unit  of  surface  most  fre- 
quently chosen  is  a  square  whose  side  is  of  unit  length.     If  the 
unit  length  is  an  inch,  the  unit  of  surface  is  a  square  whose 
side  is  an  inch.     Such  a  unit  is  called  a  square  inch.     If  the 
unit  length  is  a  foot,  the  unit  of  surface   is  a  square  whose 
side  is  a  foot,  and  the  unit  is  called  a  square  foot. 

467.  Def.     The  result  of  the  measurement  is  a  number,  which 
is  called  the  measure-number,  or  numerical  measure,  or  area  of 
the  surface. 

B  C 


A  D 

FIG.  1.    Rectangle  AC  =  15  U. 

468.  Thus,  if  the  square  U  is  contained  in  the  rectangle 
ABCD  (Fig.  1)  15  times,  then  the  measure-number  or  area  of 
rectangle  ABCD,  in  terms  of  U,  is  15.  If  the  given  square  is 
not  contained  in  the  given  rectangle  an  integral  number  of 

209 


210 


PLANE   GEOMETRY 
G 


E  H 

FIG.  2.    Rectangle  EG  =  8U+. 

times  without  a  remainder  (see  Fig.  2),  then  by  taking  a  square 
which  is  an  aliquot  part  of  U,  as  one  fourth  of  u,  and  applying 


M 


E 


FIG.  3.    Rectangle  EG  = 


H 

U+=  11J  U+. 


it  as  a  measure  to  the  rectangle  (see  Fig.  3)  a  number  will 
be  obtained  which,  divided  by  four,*  will  give  another  (and 

G 


FIG.  4.    Rectangle  A  G  =  *?$ >  £7+  =  111  U+ . 

usually  closer)  approximate  area  of  the  given  rectangle.  By 
proceeding  in  this  way  (see  Fig.  4),  closer  and  closer  approxi- 
mations of  the  true  area  may  be  obtained. 


*  It  takes  four  of  the  small  squares  to  make  the  unit  itself. 


BOOK  IV  211 

469.  If  the  sides  of   the  given   rectangle  and  of   the  unit 
square  are  commensurable,  a  square  may  be  found  which  is  an 
aliquot  part  of  U,  and  which  is  contained  in  the  rectangle  also 
an  integral  number  of  times. 

470.  If  the  sides  of   the  given  rectangle  and  of  the  unit 
square  are  incommensurable,  then  closer  and  closer  approxima- 
tions to  the  area  may  be  obtained,  but  no  square  which  is  an 
aliquot  part  of  U  will  be  also  an  aliquot  part  of  the  rectangle 
(by  definition  of  incommensurable  magnitudes).     There  is,  how- 
ever, a  definite  limit  which  is  approached  more  and  more  closely 
by  the  approximations  obtained  by  using  smaller  and  smaller 
subdivisions  of  the  unit  square,  as  these  subdivisions  approach 
zero  as  a  limit.* 

471.  Def.     The  measure-number,  or  area,  of  a  rectangle  which 
is  incommensurable  with  the  chosen  unit  square  is  the  limit 
which    successive  approximate   measure-numbers   of   the   rec- 
tangle approach  as  the  subdivisions  of  the  unit  square  approach 
zero  as  a  limit. 

For  brevity  the  expression,  the  area  of  a  figure,  is  used  to 
mean  the  measure-number  of  the  surface  of  the  figure  with 
respect  to  a  chosen  unit. 

472.  Def.     The  ratio  of  any  two  surfaces  is  the  ratio  of  their 
measure-numbers  (based  on  the  same  unit). 

473.  Def.     Equivalent   figures   are   figures  which   have   the 
same  area. 

The  student  should  note  that : 

Equal  figures  have  the  same  shape  and  size;  such  figures  can 
be  made  to  coincide. 

Similar  figures  have  the  same  shape. 
Equivalent  figures  have  the  same  size. 


Ex.  799.     Draw  two  equivalent  figures  that  are  not  equal. 

*  For  none  of  these  approximations  can  exceed  a  certain  fixed  number,  for  example 
(h  +  \)(b  +  \),  where  the  measure  applied  is  contained  in  the  altitude  h  times  with  a  re- 
mainder less  than  the  measure,  and  in  the  base  b  times  with  a  remainder  less  than  the 
measure. 


212 


PLANE   GEOMETRY 


Ex.  800.  Draw  two  equal  figures  on  the  blackboard  or  cut  them 
out  of  paper,  and  show  that  equal  figures  may  be  added  to  them  in  such 
a  way  that  the  resulting  figures  are  not  equal.  Are  they  the  same  size  ? 

Ex.  801.  Draw  figures  to  show  that  axioms  2,  7,  and  8,  when  applied 
to  equal  figures,  do  not  give  results  which  satisfy  the  test  for  equal  figures. 


s 


ni 


FIG.  1. 


FIG.  2. 


Ex.  802.  Fig.  1  above  represents  a  card  containing  64  small  squares, 
cut  into  four  pieces,  I,  II,  III,  and  IV.  Fig.  2  represents  these  four 
pieces  placed  together  in  different  positions  forming,  as  it  would  seem,  a 
rectangle  containing  65  of  these  small  squares.  By  your  knowledge  of 
similar  triangles,  try  to  explain  the  fallacy  in  the  construction. 


474.  Historical  Note.  Geometry  is  supposed  to  have  had  its  origin 
in  land  surveying,  and  the  earliest  traditions  state  that  it  had  its  beginning 
in  Egypt  and  Babylon.  The  records  of  Babylon  were  made  on  clay 
tablets,  and  give  methods  for  finding  the  approximate  areas  of  several 
rectilinear  figures,  and  also  of  the  circle.  The  Egyptian  records  were 
made  on  papyrus.  Herodotus  states  that  the,  fact  that  the  inundations  of 
the  Nile  caused  changes  in  the  amount  of  taxable  land,  rendered  it  neces- 
sary to  devise  accurate  land  measurements. 

This  work  was  done  by  the  Egyptian  priests,  and  the  earliest  manu- 
script extant  is  that  of  Ahmes,  who  lived  about  1700  B.C.  This  manu- 
script, known  as  the  Rhind  papyrus,  is  preserved  in  the  British  Museum. 
It  is  called  "Directions  for  knowing  all  dark  things,"  and  is  thought  to 
be  a  copy  of  an  older  manuscript,  dating  about  3400  B.C.  In  addition  to 
problems  in  arithmetic  it  contains  a  discussion  of  areas.  Problems  on 
pyramids  follow,  which  show  some  knowledge  of  the  properties  of  similar 
figures  and  of  trigonometry,  and  which  give  dimensions,  agreeing  closely 
with  those  of  the  great  pyramids  of  Egypt. 

The  geometry  of  the  Egyptians  was  concrete  and  practical,  unlike  that 
of  the  Greeks,  which  was  logical  and  deductive,  even  from  its  beginning. 


BOOK  IV 


213 


PROPOSITION  I.     THEOREM 

475.   The  area  of  a  rectangle  is  equal  to  the  product  of 
its  base  and  its  altitude.    (See  §  476.) 

B  C 


A  D 

Given  rectangle  ABCD,  with  base  AD  and  altitude  AB,  and  let 
U  be  the  chosen  unit  of  surface,  whose  side  is  u. 

To  prove   the  area  of  ABCD  =  AD  •  AB. 

I.    If  AD  and  AB  are  each  commensurable  with  u. 

(a)  Suppose  that  u  is  contained  in  AD  and  AB  each  an  in- 
tegral number  of  times. 


ARGUMENT 

1.  Lay  off  u  upon  AD  and  AB,  respectively.  - 

Suppose  that  u  is  contained  in  AD  r 
times,  and  in  AB  s  times. 

2.  At  the  points  of  division  on  AD  and  on  AB 

erect  J§  to  AD  and  AB,  respectively. 

3.  Then   rectangle  A  BCD  is   divided  into 

unit  squares. 

4.  There  are  r  of  these  unit  squares  in  a 

row   on   AD,   and   s  rows    of    these 
squares  in  rectangle  A  BCD. 

5.  .-.  the  area  of  ABCD  =  r  •«. 

6.  But  r  and  s  are  the  measure-numbers 

of  AD  and  AB,  respectively,  referred 
to  the  linear  unit  u. 

7.  .*.  the  area  of  ABCD  =  AD-AB,       Q.E.D. 


REASONS 

1.  §  335. 

2.  §  63. 

3.  §  466. 

4.  Arg.  1. 


5.  §  467. 

6.  Arg.  1. 


7.    §  309. 


214 


PLANE   GEOMETRY 


(6)  If  u  is  not  a  measure  of  AD  and  AB,  respectively,  but 
if  some  aliquot  part  of  u  is  such  a  measure. 
The  proof  is  left  to  the  student. 

B  C 


M 


A  QD 

II.   If  AD  and  AB  are  each  incommensurable  with  u. 


ARGUMENT 

1.  Let  m  be  a  measure  of  u.     Apply  m  as 

a  measure  to  AD  and  AB  as  many 
times  as  possible.  There  will  be  a 
remainder,  as  QD,  on  AD,  and  a  re- 
mainder, as  PB,  on  AB,  each  less 
than  m. 

2.  Through  Q  draw  QM  _L  AD,  and  through 

P  draw  PM  _L  AB. 

3.  Now  .4  Q  and  4P  are  each  commensu- 

rable with  the  measure  m,  and  hence 
commensurable  with  u. 

4.  .-.  the  area  of  rectangle  APM Q= A  Q  •  AP. 

5.  Now  take  a  smaller  measure  of  u.     No 

matter  how  small  a  measure  of  u  is 
taken,  when  it  is  applied  as  a  measure 
to  AD  and  .4Z?,  the  remainders,  QD  and 
PJ5,  will  be  smaller  than  the  measure 
taken. 

6.  .-.  the  difference  between  A  Q  and  AD 

may  be  made  to  become  and  remain 
less  than  any  previously  assigned 
segment,  however  small. 


REASONS 
1.   §  339. 


2.  §  63. 

3.  §  337. 


4.  §  475,  I. 

5.  §  335. 


6.   Arg.  5. 


BOOK  IV 


215 


ARGUMENT 

7.  Likewise    the    difference  between  AP 

and  AB  may  be  made  to  become  and 
remain  less  than  any  previously  as- 
signed segment,  however  small. 

8.  .•.  A  Q  approaches  AD  as  a  limit,  and  AP 

approaches  AB  as  a  limit. 

9.  .•.   AQ'AP    approaches    AD-AB    as  a 

limit. 

10.  Again,  the  difference  between  APMQ 

and  A  BCD  may  be  made  to  become 
and  remain  less  than  any  previously 
assigned  area,  however  small. 

11.  .*.  APMQ  approaches  ABCD  as  a  limit. 

12.  But  the  area  of  APMQ  is  always  equal 

to  AQ  -  AP. 

13.  .-.  the  area  of  ABCD  —  AD-AB. 

Q.E.D. 


9. 
10. 


11. 
12. 


REASONS 
Arg.  5. 


§  349. 
§  477. 
Arg.  5. 


§349. 
Arg.  4. 


13.    §  355. 


III.  If  AD  is  commensurable  with  u  but  AB  incommensu- 
rable with  u.  The  proof  is  left  as  an  exercise  for  the  student. 

476.  Note.     By  the  product  of  two  lines  is  meant  the  product  of 
the  measure-numbers  of  the  lines.     The  proof  that  to  every  straight  line 
segment  there  belongs  a  measure-number  is  given  in  §  595. 

477.  If  each  of  any  finite  number  of  variables  approaches  a 
finite  limit,  not  zero,  then  the  limit  of  their  product  is  equal  to 
the  product  of  their  limits.     (See  §  593.) 

478.  Cor.  I.    The  area  of  a  square  is  equal  to  the  square 
of  its  side. 

479.  Cor.  II.    Any  two  rectangles  are  to  each  other  as 
the  products  of  their  bases  and  their  altitudes. 

OUTLINE  OF  PROOF.  Denote  the  two  rectangles  by  R  and 
R',  their  bases  by  b  and  b',  and  their  altitudes  by  h  and  h',  re- 
spectively. Then  R  =  b-h  and  R'  =  b'-  h'.  .-.  —f  =  -^A. 


216  PLANE   GEOMETRY 

480.  Cor.  III.  (a)  Two  rectangles  having  equal  bases  are 
to  each  other  as  their  altitudes,  and  (b)  two  rectangles 
having  equal  altitudes  are  to  each  other  as  their  bases. 

OUTLINE  OF  PROOF 

s    R        b  •  h       h  v    R        b  •  h       b 


Ex.  803.  Draw  a  rectangle  whose  base  is  7  units  and  whose  altitude 
is  4  units  and  show  how  many  unit  squares  it  contains. 

Ex.  804.  Find  the  area  of  a  rectangle  whose  base  is  12  inches  and 
whose  altitude  is  5  inches. 

Ex.  805.  Find  the  area  of  a  rectangle  whose  diagonal  is  10  inches, 
and  one  of  whose  sides  is  6  inches. 

Ex.  806.  If  the  area  of  a  rectangle  is  60  square  feet,  and  the  base, 
5  inches,  what  is  the  altitude  ? 

Ex.  807.  If  the  base  and  altitude  of  a  rectangle  are  2J  inches  and  l\ 
inches,  respectively,  find  the  area  of  the  rectangle. 

Ex.  808.     Find  the  area  of  a  square  whose  diagonal  is  8  \/2  inches. 

Ex.  809.  Find  the  successive  approximations  to  the  area  of  a  rec- 
tangle if  its  sides  are  V  10  and  V  5,  respectively,  using  3  times  2  for  the 
first  approximation,  taking  the  square  roots  to  tenths  for  the  next,  to 
hundredths  for  the  next,  etc. 

Ex.  810.  Compare  two  rectangles  if  a  diagonal  and  a  side  of  one  are 
d  and  s,  respectively,  while  a  diagonal  and  side  of  the  other  are  d'  and  s1. 

Ex.  811.  Construct  a  rectangle  whose  area  shall  be  three  times  that 
of  a  given  rectangle. 

Ex.  812.  Construct  a  rectangle  which  shall  be  to  a  given  rectangle  in 
the  ratio  of  two  given  lines,  m  and  n. 

Ex.  813.  Compare  two  rectangles  whose  altitudes  are  equal,  but 
whose  bases  are  15  inches  and  3  inches,  respectively. 

Ex.  814.  From  a  given  rectangle  cut  off  a  rectangle  whose  area  is 
two  thirds  that  of  the  given  one. 

Ex.  815.  If  the  base  and  altitude  of  a  certain  rectangle  are  12  inches 
and  8  inches,  respectively,  and  the  base  and  altitude  of  a  s°cond  rectangle 
are  6  inches  and  4  inches,  respectively,  compare  their  areas. 


BOOK   IV 


217 


PROPOSITION  II.     THEOUEM 

481.   The  area  of  a  parallelogram  equals  the  product  of 
its  base  and  its  altitude. 

R  b C 


A  E 

Given   HH  ARCD,  with  base  6  and  altitude  h. 
To  prove   area  of  A  ROD,  =  b  -  h. 

ARGUMENT 

1.  Draw   the  rectangle  ERCF,  having  b  as 

base  and  h  as  altitude. 

2.  In  rt.  A  DCF  and  ARE,  DC  =  AR. 

Also  CF  =  RE. 
.-.  A  DCF  =  A  ARE. 
Now  figure  ARCF  =  ngure^flC^. 

.-.  area  of  ARCD  =  area  of  ERCF. 
But  area  of  ERCF  =  b  -  h. 
.-.  area  of  ARCD  =  b  -  h.      Q.E.D. 


REASONS 
1.    §223. 


§232. 
§232. 
§211. 
By  iden. 
§  54,  3. 
§475. 
§  54,  1. 


482.  Cor.  I.     Parallelograms   having  equal   bases  and 
equal  altitudes  are  equivalent. 

483.  Cor.  n.     Any  two  parallelograms  are  to  each  other 
as  the  products  of  their  bases  and  their  altitudes. 

HINT.     Give  a  proof  similar  to  that  of  §  479. 

484.  Cor.  III.     (a)    Two  parallelograms  having  equal 
bases  are  to  each  other  as  their  altitudes,  and  (b)  two  par- 
allelograms having  equal  altitudes  are  to  each  other  as 
their  bases.     (HINT.     Give  a  proof  similar  to  that  of  §  480.) 


218 


PLANE   GEOMETRY 


PROPOSITION  III.     THEOREM 

485.   Tlw  area  of  a  triangle  equals  one  half  the  product 
of  its  base  and  its  altitude. 


1. 


REASONS 

CB 

,    and 

1. 

§179. 

CA. 

Let 

2. 

§220. 

fJ. 

3. 

§236. 

4. 

§481. 

Q.E.I). 

o. 

§54,  8  a. 

A  b  C 

Given  A  ABC,  with  base  b  and  altitude  h. 
To  prove   area  of  A  ABC  =  ^b-  h. 

ARGUMENT 

Through  A  draw  a  line 
through  B  draw  a  line 
these  lines  intersect  at  X. 

2.  Then  JAr£(7  is  a  O. 

3.  /.  A 

4.  But  area  of 

5.  .-.  area  of  A 

486.  Cor.  I.     Triangles  having  equal  bases  and  equal 
altitudes  are  equivalent. 

487.  Cor.  II.     Any  two  triangles  arc  to  each  other  as  the 
products  of  their  bases  and  their  altitudes. 

OUTLINE  OF  PROOF 

Denote  the  two  A  by  T  and  T',  their  bases  by  b  and  b',  and 
their  altitudes  by  h  and  h',  respectively.     Then  T  =  i  b  •  h  and 

'b^h1' 

488.  Cor.  III.     (a)  Two  triangles  having  equal  bases  are 
to  each  other  as  their   altitudes,  and  (b)  two   triangles 
having  equal  altitudes  are  to  each  as  their  bases. 


BOOK  IV  219 

OUTLINE  OF  PROOF 


?~b.h'Ji  T'b'.hb' 

489.   Cor.  IV.     A  triangle  is  equivalent  to  one  half  of 
a  parallelogram  having  the  same  base  and  altitude. 


Ex.  816.     Draw  four  equivalent  parallelograms  on  the  same  base. 

Ex.  817.  Find  the  area  of  a  parallelogram  having  two  sides  8  inches 
and  12  inches,  respectively,  and  the  included  angle  60°.  Find  the  area 
if  the  included  angle  is  45°. 

Ex.  818.  Find  the  ratio  of  two  rhombuses  whose  perimeters  are  24 
inches  and  16  inches,  respectively,  and  whose  smaller  base  angles  are  30°. 

Ex.  819.  Find  the  area  of  an  equilateral  triangle  having  a  side  equal 
to  6  inches. 

Ex.  820.  Find  the  area  of  an  equilateral  triangle  whose  altitude  is 
8  inches. 

Ex.  821.    Construct  three  or  more  equivalent  triangles  on  the  same  base. 

Ex.  822.  Find  the  locus  of  the  vertices  of  all  triangles  equivalent  to 
a  given  triangle  and  standing  on  the  same  base. 

Ex.  823.  Construct  a  triangle  equivalent  to  a  given  triangle  and  hav- 
ing one  of  its  sides  equal  to  a  given  line. 

Ex.  824.  Construct  a  triangle  equivalent  to  a  given  triangle  and  hav- 
ing one  of  its  angles  equal  to  a  given  angle. 

Ex.  825.  Construct  a  triangle  equivalent  to  a  given  triangle  and  hav- 
ing two  of  its  sides  equal,  respectively,  to  two  given  lines. 

Ex.  826.  Divide  a  triangle  into  three  equivalent  triangles  by  drawing 
lines  through  one  of  its  vertices. 

Ex.  827.  Construct  a  triangle  equivalent  to  f  of  a  given  triangle  ; 
|  of  a  given  triangle. 

Ex.  828.     Construct  a  triangle  equivalent  to  a  given  square. 

Ex.  829.  The  area  of  a  rhombus  is  equal  to  one  half  the  product  of 
its  diagonals. 

Ex.  830.  If  from  any  point  in  a  diagonal  of  a  parallelogram  lines  are 
drawn  to  the  opposite  vertices,  two  pairs  of  equivalent  triangles  are  formed. 

Ex.  831.  Two  lines  joining  the  mid-point  of  the  diagonal  of  a  quadri- 
lateral to  the  opposite  vertices  divide  the  figure  into  two  equivalent  parts. 

Ex.  832.  Find  the  area  of  a  triangle  if  two  of  its  sides  are  6  inches 
and  9  inches,  respectively,  and  the.  included  angle  is  60°. 


220 


PLANE   GEOMETRY 


PROPOSITION  IV.     PROBLEM 

490.   To  derive  a  formula  for  the  area  of  a  triangle  in 
terms  of  its  sides. 


C 


B 


Given  A  ABC,  with  sides  ft,  6,  and  c. 

To  derive   a  formula  for  the  area  of  A  ABC  in  terras  of  a,  6, 

and  c. 

ARGUMENT  KEASONS 

1.    Let  ha  denote  the  altitude  upon  «,  p  the  pro-     1.  §  485. 
jection  of   b.  upon  ft,  and    T  the  area  of 
A  ABC. 


Then  To 

L 

2.   h*=V-p>  = 
3. 


(Fig.  2). 


Va  .  '2a 

_(ft-|-  6  +  c)(ft-f-  b— c)(c-\-a—  6)(c— ft 
4ft2 

^  /(g-f-64.c)(ft-f6-c)(c+ft-'^)(c-ft 

4ft2 
6.  Now  let  ft  +  &  -f-  c  =  2  s. 

Then       a  +  b  —  c  =  2s  —  2c  =  2(s  —  c"); 


2.  §  447. 

3.  §456. 

4.  §  309. 


5.  §  54,  13. 

6.  §  54,  3, 


BOOK  IV 


221 


ARGUMENT 


-6)2(.-c) 


7.ThenA.  =  >/2i2<J     f^L 

2     __^____ T 

=  \  V<«  -  a)(* -&)(«-  c). 


a     2 


—  a)(s  —  b)(s  —  c). 


Q.E.F. 


REASONS 
7.   §  309. 


8.  §  309. 


Ex.  833.     Find  the  area  of  a  triangle  whose  sides  are  7,  10,  and  13. 

Ex.  834.  If  the  sides  of  a  triangle  are  a,  &,  and  c,  write  the  formula 
for  the  altitude  upon  6  ;  upon  c.  (See  Prop.  IV,  Arg.  7. ) 

Ex.  835.  In  triangle  ABC,  a  =  8,  b  =  12,  c  =  16  ;  find  the  area  of 
triangle  ABC ;  the  altitude  upon  b  ;  the  altitude  upon  c. 


PROPOSITION  V.     THEOREM 

491.  Th<e  area  of  a  triangle  is  equal  to  one  half  the 
product  of  its  perimeter  and  the  radius  of  the  inscribed 
circle. 


Given  A  ABC,  with  area  T,  sides  o,  6,  and  c,  and  radius  of 
inscribed  circle  r. 

To  prove    T=  \  (a  +  b  +  c)  r. 

OUTLINE  OF  PROOF 

1.  Area  of  A  OBC  =  \  a  •  r;  area  of  A  OCA  =  1  b  -  r ;  area 
of  A  OAB  =  1  c  •  r.  2.  . ' .  T  =  %  (a  +  b  +  c)  r.  Q.E.D. 

492.  Cor.  The  area  of  any  polygon  circumscribed 
about  a  circle  is  equal  to  one  half  its  perimeter  multi- 
plied by  the  radius  of  the  inscribed  circle. 


222  PLANE   GEOMETRY 

Ex.  836.  If  the  area  of  a  triangle  is  15  \/3  square  inches  and  its  sides 
are  3,  5,  and  7  inches,  find  the  radius  of  the  inscribed  circle. 

Ex.  837.  Derive  a  formula  for  the  radius  of  a  circle  inscribed  in  a 
triangle  in  terms  of  the  sides  of  the  triangle. 

OUTLINE  OF  SOLUTION 
1.         r=i(a +  & +  <>)»•=  i(2*)r  =  w. 


—  a)  (a  —  ft)  < s  —  c) 

Q.E.F. 


Ex.  838.     Derive  a  formula  for  the  radius  of  a  circle  circumscribed 
about  a  triangle,  in  terms  of  the  sides  of  the  triangle. 

OUTLINE  OF  SOLUTION     (See  figure  for  Ex.  798.) 
1.    dh  =  ac;  i.e.  dor2It=—. 


h 

abc 


Q.E.F. 


2  h     4V  s(s  —a)  (s  —  6)  (s  -  c) 

Ex.  839.  If  the  sides  of  a  triangle  are  9,  10,  and  11,  find  the  radius 
of  the  inscribed  circle  ;  the  radius  of  the  circumscribed  circle. 

Ex.840.  The  sides  of  a  triangle  are  3  a,  4  a,  and  5  a.  Find  the 
radius  of  the  inscribed  circle ;  the  radius  of  the  circumscribed  circle. 
What  kind  of  a  triangle  is  it  ?  Verify  your  answer  by  comparing  the 
radius  of  the  circumscribed  circle  with  the  longest  side. 

Ex.  841.  Derive  formulas  for  the  bisectors  of  the  angles  of  a  triangle 
in  terms  of  the  sides  of  the  triangle. 

OUTLINE  OF  SOLUTION     (See  figure  for  Ex.  797.) 
1.    tb2  =  ac  —  rs.       2.    Buta:c=r:s.  3.    .•.  a  -\-  c  :  a  =b  :  r. 


4:.       . 

ac  (a 

',  +  b  +  c)(a  —  6  -f  c" 

K.ewise  «  —  —    —  •          o.    .  .  t( 

a  +  c 

(a-fc)2 

a 

)        A  (ZC6(^5         OJ             M          ^      . 

8    T, 

(«  +  c)2 
ikp.wise  t»  —  x// 

(«  +  c)2 

a  +  c 

:f>af  «       r^  . 

Q.E.F. 

Find  r,  E,  T,  ha,  wa,  and  «„,  having  given : 
Ex.  842.     a  =  11,  b  =  9,  c  =  1(5.     What  kind  of  an  angle  is  C  ? 
Ex.  843.     a  =  13,  b  =  15,  c  =  20.     What  kind  of  an  angle  is  C  ? 
Ex.  844.     a  =  24,  b  =  10,  c  =  26.     What  kind  of  an  angle  is  C  ? 


BOOK   IV 


223 


TRANSFORMATION   OF   FIGURES 

493.    Def.     To   transform   a   figure   means   to   find  another 
figure  which  is  equivalent  to  it. 


PROPOSITION  VI.     PKOBLEM 

494.     To  construct   a  triangle   equivalent  to  a  given 
polygon. 


Given    polygon  ABCDEF. 

To  construct    a  A  =0=  polygon  ABCDEF. 

(a)   Construct  a  polygon  o  ABCDEF,  but  having  one  side  less. 

I.    Construction 

1.  Join  any  two  alternate  vertices,  as  C  and  A. 

2.  Construct  BG  II  CA,  meeting  FA  prolonged  at  G.     §  188. 

3.  Draw  CG. 

4.  Polygon  GCDEF  =0=  polygon  ABCDEF  and  has  one  side  less. 


II.    Proof 

ARGUMENT 

1.  A  AGC  and  ABC  have  the  same  base 

CM,  and  the  same  altitude,  the  J_  be- 
tween the  Us  CA  and  BG. 

2.  .'.  A  AGC  o  A  ABC. 

3.  -But  polygon  A  CDEF  =  polygon  A  CDEF. 

4.  .-.  polygon  GCDEF  =0=  polygon  ABCDEF. 

Q.E.D. 


REASONS 
1.    §235. 


2.  §486. 

3.  By  iden. 

4.  §  54,  2. 


224 


PLANE    GEOMETRY 


G 


F 


H 


(b)    In  like  manner,  reduce  the  number  of  sides  of  the  new 
polygon  GGDEF  until  A  DHK  is  obtained. 

The  construction,  proof,  and  discussion  are  left  as  an  exercise 
for  the  student. 

Ex.  845.  Transform  a  scalene  triangle  into  an  isosceles  triangle. 

Ex.  846.  Transform  a  trapezoid  into  a  right  triangle. 

Ex.  847.  Transform  a  parallelogram  into  a  trapezoid. 

Ex.  848.  Transform  a  pentagon  into  an  isosceles  triangle. 

Ex.  849.  Construct  a  triangle  equivalent  to  f  of  a  given  trapezium. 

Ex.  850.  Transform  |  of  a  given  pentagon  into  a  triangle. 

Ex.  851.  Construct  a  rhomboid  and  a  rhombus  which  are  equivalent, 
and  which  have  a  common  diagonal. 


PROPOSITION  VII.     THEOREM 

495.     The  area  of  a  trapezoid  equals  the  product  of  its 
altitude  and  one  half  tfa  sum  of  its  bases. 

V 


D 


Given   trapezoid  AFED,  with  altitude  h  and  bases  b  and  b'. 
To  prove   area  of  AFED  =  |  (b  +  b')h. 


BOOK  IV 


225 


ARGUMENT 

1.  Draw  the  diagonal  AE. 

2.  The  altitude  of   A  AED,  considering  b 

as  base,  is  equal  to  the  altitude  of 
A  AFE,  considering  b'  as  base,  each 
being  equal  to  the  altitude  of  the 
trapezoid,  h. 

3.  .'.  area  of  A  AED  =  ^  b  -  h. 

4.  Area  of  A  AFE  =  *-  b'  -  h. 

5.  .-.  area  of  trapezoid  AFED  =  %(b  +  b')h. 

Q.E.D. 


REASONS 

1.  §  54,  15. 

2.  §  235. 


§485. 
§  485. 
§  54,  2. 


496.  Cor.     The  area  of  a  trapezoid  equals  the  product  of 
its  altitude  and  its  median. 

497.  Question.     The  ancient  Egyptians,  in  attempting  to  find  the 
area  of  a  field  in  the  shape  of  a  trapezoid,  multiplied  one  half  the  sum  of 
the  parallel  sides  by  one  of  the  other  sides.     For  what  figure  would  this 
method  be  correct  ?  

Ex.  852.  Find  the  area  of  a  trapezoid  whose  bases  are  7  inches  and 
9  inches,  respectively,  and  whose  altitude  is  5  inches. 

Ex.  853.  Find  the  area  of  a  trapezoid  whose  median  is  10  inches  and 
whose  altitude  is  6  inches. 

Ex.  854.  Through  a  given  point  in  one  side  of  a  given  parallelogram 
draw  a  line  which  shall  divide  the  parallelogram  into  two  equivalent  parts. 
Will  these  parts  be  equal  ? 

Ex.  855.  Through  a  given  point  within  a  parallelogram  draw  a  line 
which  shall  divide  the  parallelogram  into  two  equivalent  parts.  Will 
these  parts  be  equal  ? 

Ex.  856.  If  the  mid-point  of  one  of  the  non-parallel  sides  of  a  trape- 
zoid is  joined  to  the  extremities  of  the  other  of  the  non-parallel  sides,  the 
area  of  the  triangle  formed  is  equal  to  one  half  the  area  of  the  trapezoid. 

Ex.  857.  Find  the  area  of  a  trapezoid  whose  bases  are  b  and  b'  and 
whose  other  sides  are  each  equal  to  s. 

Ex.  858.  If  the 'sides  of  any  quadrilateral  are  bisected  and  the  points 
of  bisection  joined,  the  included  figure  will  be  a  parallelogram  equal  in 
area  to  half  the  original  figure. 


226 


PLANE   GEOMETRY 


PROPOSITION  VIII.     THEOREM 

498.  Two  triangles  which  have  an  angle  of  one  equal 
to  an  angle  of  the  other  are  to  each  other  as  tJie  products 
of  the  sides  including  the  equal  angles. 

B 


Given    A  AB  C  and  DEF,  with  Z  A  ±= 
A  ABC      AC  •  AB 


To  prove 


A  DEF      DF  •  DE 


ARGUMENT 

1.    Let  h  be  the  altitude  of   A  ABC  upon 
side  AC,  and  h'  the  altitude  of  A  DEF 
upon  side  DF.     Then, 
A  ABC       AC  •  h       AC 


REASONS 

§  487. 


A  DEF 

DF  •  h1 

DF    h1' 

2. 

3. 

In  rt.  A  AB  G  and  DEK,  /.  A  =  /.  D. 

.-.  AABG  ~  ADEK. 

2. 
3. 

By  hyp. 
§  422. 

4. 

h'       DE 

4. 

§  424,  2. 

5. 

AABC 

AC    AB 

AC  -  AB 
—                            n  TF  n 

5. 

§309. 

A  DEF 

DF    DE 

DF  -DE 

Ex.  859.  Draw  two  triangles  upon  the  blackboard,  so  that  an  angle 
of  one  shall  equal  an  angle  of  the  other.  Give  a  rough  estimate  in  inches 
of  the  sides  including  the  equal  angles  in  the  tw6  triangles,  and  compute 
the  numerical  ratio  of  the  triangles. 

Ex.  860.  If  two  triangles  have  an  angle  of  one  supplementary  to  an 
angle  of  the  other,  the  triangles  are  to  each  other  as  the  products  of  the 
sides  including  the  supplementary  angles. 


BOOK  IV 


227 


PROPOSITION  IX.     PROBLEM 

499.    To  construct  a  square  equivalent  to  a  given  tri- 
B 


angle. 


\ 


D 


S 

/ 
/ 

/ 

/ 

\ 
V 

/ 

Given   A  ABC,  with  base  b  and  altitude  h. 
To  construct   a  square  =c=  A  AB  C. 

I.    Analysis 

1.  Let  #  =  the  side  of  the  required  square;  then  #2 
of  required  square. 

2.  i  b  •  h  —  area  of  the  given  A  ABC. 

3.  .'.a?  =  ±b-h. 

4.  /.  J -b  :  x  =  x  :  h. 

5.  .-.  the  side  of  the  required  square  will  be  a  mean  pro- 
portional between  |-  b  and  h. 

II.    Construction 

1.  Construct  a  mean  proportional  between  7]  b  and  h.     Call 
it  x.     §  445. 

2.  On  a?,  as  base,  construct  a  square,  S. 

3.  S  is  the  required  square. 


III.    Proof 


ARGUMENT 

1. 

l  6  :  x  =  x  :  h. 

o 

2         177 

w. 

.  .  X    —  -Q  Oil. 

3.   But 

x2  =  area  of  S. 

4.    And 

^  b  •  h  =  area  of  A  ABC. 

5. 

.'.  So  A  ABC. 

Q.E.D. 


REASONS 

1.  By  cons. 

2.  §  388. 

3.  §  478. 

4.  §485. 

5.  §  54,  1. 


228  PLANE   GEOMETRY 

IV.    The  discussion  is  left  as  an  exercise  for  the  student. 


500.  Question.     Could  x  be  constructed  as  a  mean  proportional  be- 
tween b  and  £  h ? 

501.  Problem.     To  construct  a  square  equivalent  to  a 
given  parallelogram. 

Ex.  861.     Construct  a  square  equivalent  to  a  given  rectangle. 
Ex.  862.     Construct  a  square  equivalent  to  a  given  trapezoid. 
Ex.  863.     Upon  a  given  base  construct  a  triangle  equivalent  to  a 
given  parallelogram. 

Ex.  864.  Construct  a  rectangle  having  a  given  base  and  equivalent 
to  a  given  square.  

502.  Props.  VI,  VIII,  and  IX  form  the  basis  of  a  large 
class  of  important  constructions. 

(a)  Prop.  VI  enables  us  to  construct  a  triangle  equivalent 
to  any  polygon.  It  is  then  an  easy  matter  to  construct  a 
trapezoid,  an  isosceles  trapezoid,  a  parallelogram,  a  rectangle, 
or  a  rhombus  equivalent  to  the  triangle  and  hence  equivalent 
to  the  given  polygon. 

(6)  Prop.  VIII  gives  us  a  method  for  constructing  an  equi- 
lateral triangle  equivalent  to  any  given  triangle.  (See  Ex.  865.) 
Hence  Prop.  VIII,  with  Prop.  VI,  enables  us  to  construct  an 
equilateral  triangle  equivalent  to  any  given  polygon. 

(c)  Likewise  Prop.  IX,  with  Prop.  VI,  enables  us  to  con- 
struct a  square  equivalent  to  any  given  polygon  or  to  any 
fractional  part  or  to  any  multiple  of  any  given  polygon. 


Ex.  865.  (a)  Transform  triangle  ABC  into  triangle  DBG,  retaining 
base  BC  and  making  angle  DBC=  60°. 

(6)  Transform  triangle  DBC  into  triangle  EBF,  retaining  angle 
DBC  —  60°  and  making  sides  EB  and  BF  equal.  (Each  will  be  a  mean 
proportional  between  DB  and  BC.} 

(c)  What  kind  of  a  triangle  is  EBF? 

Ex.  866.     Transform  a  parallelogram  into  an  equilateral  triangle. 

Ex.  867.  Construct  an  equilateral  triangle  equivalent  to  f  of  a  given 
trapezium. 


BOOK   IV 


229 


Ex.  868.  Construct  each  of  the  following  figures  equivalent  to  f  of  a 
given  irregular  pentagon  :  (1)  a  triangle;  (2)  an  isosceles  triangle  ;  (3)  a 
right  triangle  ;  (4)  an  equilateral  triangle  ;  (5)  a  trapezium ;  (6)  a  trape- 
zoid;  (7)  an  isosceles  trapezoid  ;  (8)  a  parallelogram;  (9)  a  rhombus; 
(10)  a  rectangle  ;  (11)  a  square. 

Ex.  869.  Transform  a  trapezoid  into  a  right  triangle  having  the 
hypotenuse  equal  to  a  given  line.  What  restrictions  are  there  upon  the 
given  line  ? 

Ex.  870.  Construct  a  triangle  equivalent  to  a  given  trapezoid,  and 
having  a  given  line  as  base  and  a  given  angle  adjacent  to  the  base. 


PROPOSITION  X.     THEOREM 

503.   Two  similar  triangles  are  to  each   other  as  the 
squares  of  any  two  homologous  sides. 

B 


b1 


Given   two  similar  A  ABC  and  A'B'C',  with,  b  and  b'  two 
homol.  sides. 

A  ABC       b2 


To  prove 


A  A'B'C'      b'2 


ARGUMENT 
1.    A  ABC  ~  A  A'B'C1. 

.  A   ABC        b  -  c 


.>. 

X  lltUl      —  

A  A'B'C'     b1  • 

c'     b1    c' 

4. 

But                c-  =  ir 

c'      6' 

K 

A  ABC       b 

b       b2 

'  AA'B'C'~b' 

b1  ~  b'2' 

REASONS 

1.  By  hyp. 

2.  §  424,  1. 

3.  §  498. 

4.  §  424,  2. 

5.  §  309. 


230  PLANE  GEOMETRY 

504.   Cor.    Two  similar  triangles  are  to  each  other  as  the 
squares  of  any  two  homologous  altitudes.     (See  §  435.) 


Ex.  871.  Two  similar  triangles  are  to  each  other  as  the  squares  of 
two  homologous  medians. 

Ex.  872.  Construct  a  triangle  similar  to  a  given  triangle  and  having 
an  area  four  times  as  great. 

Ex.  873.  Construct  a  triangle  similar  to  a  given  triangle  and  having 
an  area  twice  as  great. 

Ex.  874.  Divide  a  given  triangle  into  two  equivalent  parts  by  a  line 
parallel  to  the  base. 

Ex.  875.     Prove  Prop.  X  by  using  §  487. 

Ex.  876.  Draw  a  line  parallel  to  the  base  of  a  triangle  and  cutting  off 
a  triangle  that  shall  be  equivalent  to  one  third  of  the  given  triangle. 

Ex.  877.  In  two  similar  triangles  a  pair  of  homologous  sides  are  10 
feet  and  6  feet,  respectively.  Find  the  homologous  side  of  a  similar  tri- 
angle equivalent  to  their  difference. 

-  Ex.  878.     Construct  an  equilateral  triangle  whose  area  shall  be  three 
fourths  that  of  a  given  square. 

PROPOSITION  XI.     THEOREM  \ 

505.  Two  similar  polygons  are  to  each  other  as  the 
squares  of  any  two  homologous  sides. 


Given  two  similar  polygons  P  and  Pf  in  which  a  and  a',  b 
and  b',  etc.,  are  pairs  of  homol.  sides. 

P      a2 
To  prove  -=-. 


BOOK  IV 


231 


ARGUMENT 

1.  Draw  all  possible  diagonals  from  any 

two  homol.  vertices,  as  y  and  V'. 

2.  Then  the  polygons  Will  be  divided  into 

the  same  number  of  A  ^  each  to 
each  and  similarly  placed,  as  A 1  and 
A  I',  A  II  and  A  II',  etc. 

3.  Then 


AT      a'2 


•*. 

Air 

e'2' 

A  III 

_d2 

5. 

Aiir 

d'2 

6.    But 

a 

_  e 

d_ 

I1' 

=  (*L- 

A  I       A  II       A  III 


a'       e* 


9.   .'. 


10. 


A  I'      AH'      A  III' 

A  I  +  A  II  -f  A  III  =  A  I  _  a2 

A  r  +  A  ir  +  A  in'  ~~  A  i'  ~~  ^2' 


REASONS 

1.  §  54,  15. 

2.  §  439. 

3.  §  503. 

4.  §  503. 

5.  §  503. 

6.  §  419. 

7.  §54,13. 

8.  §54,1. 

9.  §  401. 
10.  §  309. 


506.    Cor.    Two  similar  polygons  are  to  each  other  as  the 
squares  of  any  two  homologous  diagonals. 


Ex.  879.     If  one  square  is  double  another,  what  is  the  ratio  of  their 
sides  ? 

Ex.  880.     Divide  a  given  hexagon  into  two  equivalent  parts  so  that 
one  part  shall  be  a  hexagon  similar  to  the  given  hexagon. 

Ex.  881.     The  areas  of  two  similar  rhombuses  are  to  each  other  as 
the  squares  of  th^ir  homologous  diagonals. 

Ex.  882.     One  side  of  a  polygon  is  8  and  its  area  is  120.     The  homol- 
ogous side  of  a  similar  polygon  is  12  ;  find  its  aiva. 


232 


PLANE   GEOMETRY 


PROPOSITION  XII,     THEOREM 

507.  Tlw  square  described  on  the  hypotenuse  of  a 
right  triangle  is  equivalent  to  the  sum  of  the  squares 
described  on  the  otlwr  two  sides. 


E 


Given   rt.  A  ABC,  right-angled  at  C,  and  the  squares  described 
on  its  three  sides. 

To  prove    square  AD  =0=  square  BF  +  square  CH. 


ARGUMENT 

1.  From  C  draw  CM  J_  AB,  cutting  AB  at  L 

and  KD  at  M. 

2.  Draw  CK  and  BH. 

3.  A  ACG,  BCA,  and  FOB  are  all  rt.  A. 

4.  .'.  ACF  and  GCB  are  str.  lines. 

5.  Ill  A  CAK  and  HAS,  CA  =  HA,  AK  =  AB. 

6.  Z  CAB  =  Z  CAB. 


8.  .-.  ZCAK= 

9.  /.  ACAK  =  AHAB. 

10.  A  GMiTand  rectangle  AM  have  the  same 
base  AK  and  the  same  altitude,  the 
JL  between  the  Us  AK  and  Cy3f. 


REASONS 

1.  §155. 

2.  §  54,  15. 

3.  By  hyp. 

4.  §76. 

5.  §  233. 

6.  By  iden. 

7.  §64. 

8.  §  54,  2. 

9.  §107. 
10.  §235. 


BOOK   IV 


233 


ARGUMENT 

11.  /.A  CAK  o=  1  rectangle  AM. 

12.  Likewise  A  HAB  and  square  CH  have 

the  same  base  HA  and  the  same  alti- 
tude, the  _L  between  Us  HA  and  GB. 

13.  .'.A HAB  =0= 1  square  Ctf, 

14.  But  A  CAK  —  A  #J£. 

15.  .'.  £  rectangle  ^13f=c=  ^  square  (7#. 

16.  .'.  rectangle  AM =0  square  CH. 

17.  Likewise,  by  drawing   CD  and  AE,  it 

may  be  proved  that  rectangle  LD  =c= 
square  .W. 

18.  .*.  rectangle  JJf-f-  rectangle  LDo  square 

CH  -h  square  BF. 

19.  .'.  square  AD=c=  square  CH+  square  BF. 

Q.E.D. 


REASONS 

11.  §489. 

12.  §235. 


13.  §489. 

14.  Arg.  9. 

15.  §  54,  1. 

16.  §54,  7  a. 

17.  By  steps  simi- 

lar to  5-16. 

18.  §54,2. 

19.  §309. 


508.  Cor.  I.     The  square  described  on  either  side  of  a 
right  triangle  is  equivalent  to  the  square  described  on 
the  hypotenuse  minus  the  square  described  on  tlie  other 
side. 

509.  Cor.  n.     If  similar  polygons  are  described  on  the 
three  sides  of  a  right  triangle  as  homologous  sides,  the 
polygon   described,   on   the   hypotenuse   is   equivalent   to 
the  sum  of  the  polygons  described  on  the  other  two  sides. 

Given   rt.  A  ABC,  right-angled  at  C,  and  let  P,  Q,  and  R  be  ~ 
polygons  described  on  a,  b,  and  c,  respectively,  as  hornol.  sides. 
To  prove    R  o  P  -f  Q. 

ARGUMENT   ONLY 

2.   ^  =  ^. 


3. 


R       c2 
P+  Q 


c1 


P-hQ. 


Q.E.D. 


Ex.  883.     The  square  on  the  hypotenuse  of  an  isosceles  right  triangle 
is  equivalent  to  four  times  the  triangle. 


234 


PLANE   GEOMETRY 


510.  Historical  Note.  Prop.  XII  is  usually  known  as  the  Pythago- 
rean Proposition,  because  it  was  discovered  by  Pythagoras.  The  proof 
given  here  is  that  of  Euclid  (about  300  B.C.). 

Pythagoras  (569-500  B.C.), 
one  of  the  most  famous  mathe- 
maticians of  antiquity,  was 
born  at  Samos.  He  spent  his 
early  years  of  manhood  study- 
ing under  Thales  and  traveled 
in  Asia  Minor  and  Egypt  and 
probably  also  in  Babylon  and 
India:  He  returned  to  Samos 
where  he  established  a  school 
that  was  not  a  great  success. 
Later  he  went  to  Crotona  in 
Southern  Italy  and  there 
gained  many  adherents.  He 
formed,  with  his  closest  fol- 
lowers, a  secret  society,  the 
members  of  which  possessed  PYTHAGORAS 

all  things  in  common.     They 

used  as  their  badge  the  five-pointed  star  or  pentagram  which  they  knew 
how  to  construct  and  which  they  considered  symbolical  of  health.  They 
ate  simple  food  and  practiced  severe  discipline,  having  obedience,  temper- 
ance, and  purity  as  their  ideals.  The  brotherhood  regarded  their  leader 
with  reverent  esteem  and  attributed  to  him  their  most  important  dis- 
coveries, many  of  which  were  kept  secret. 

Pythagoras  knew  something  of  incommensurable  numbers  and  proved 
that  the  diagonal  and  the  side  of  a  square  are  incommensurable. 

The  first  man  who  propounded  a  theory  of  incommensurables  is  said  to 
have  suffered  shipwreck  on  account  of  the  sacrilege,  since  such  numbers 
were  thought  to  be  symbolical  of  the  Deity. 

Pythagoras,  having  incurred  the  hatred  of 
his  political  opponents,  was  murdered  by 
them,  but  his  school  was  reestablished  after 
his  death  and  it  flourished  for  over  a  hundred 
years.  

Ex.  884.  Use  the  adjoining  figure  to 
prove  the  Pythagorean  theorem. 

Ex.  885.  Construct  a  triangle  equivalent 
to  the  sum  of  two  given  triangles.  B 


BOOK  IV 


235 


Ex.  886.     The  figure  represents  a  farm  drawn  to  the  scale  indicated. 
Make    accurate    measure- 
ments and   calculate     ap- 
proximately   the    number 
of  acres  in  the  farm. 

Ex.  887.  A  farm 
XYZW,  in  the  form  of  a 
trapezium,  has  the  follow- 
ing dimensions  :  XY  =  60 
rods,  YZ  =  70  rods,  ZW 
=  90  rods,  WX  =  100  rods, 
and  XZ  =  66  rods.  Draw 
a  plot  of  the  farm  to  the 


i.*                    F~~  —  1 

1  1 

calculate   the   area  of   the       Q      20     40 
farm  in  acres. 

80 

160 

511.    Def.    By  the  rectangle  of  two  lines  is 

meant  the  rectangle  having  these  two  lines 
as  adjacent  sides. 

a2 

ab 

ab 

62 

Ex.  888.     The  square  described  on  the  sum 
two  lines  is  equivalent  to  the  sum  of  the  squa 
described  on  the  lines  plus  twice  their  rectangl 

Ex.  889.     The  square   described  on    the 
difference  of  two  lines  is  equivalent  to  the 
sum   of  the  squares  described  on  the  lines 
diminished  by  twice  their  rectangle. 
HINT.     Let  AB  and  CB  be  the  given  lines. 

of 
res 

e. 

a            b 

Ex.  890.  The  rectangle  whose  sides  are  the 
sum  and  difference  respectively  of  two  lines  is 
equivalent  to  the  difference  of  the  squares  described 
on  the  lines. 

HINT.     Let  AB  and  BC  be  the  given  lines. 


B     C 


Ex.  891.     Write  the  three  algebraic  formulas  corresponding  to  the 
last  three  exercises. 


236 


PLANE   GEOMETRY 


PROPOSITION  XIII.     PROBLEM 

512.   To  construct  a  square  equivalent  to  the  sum  of 
two  given  squares. 


Given   squares  P  and  Q. 

To  construct   a  square  =0=  the  sum  of  P  and  Q. 

I.    Construction 

1.  Construct  the  rt.   A  ABC,  having 'for  its  sides  p  and  q, 
the  sides  of  the  given  squares. 

2.  On  r,  the  hypotenuse  of  the  A,  construct  the  square  R. 

3.  R  is  the  required  square. 

II.     The  proof  and  discussion  are  left  to  the  student. 


Ex.  892.  Construct  a  square  equivalent  to  the  sum  of  three  or  more 
given  squares. 

Ex.  893 .    Construct  a  square  equivalent  to  the  difference  of  two  squares. 

Ex.  894.  Construct  a  square  equivalent  to  the  sum  of  a  given  square 
and  a  given  triangle. 

Ex.  895.  Construct  a  polygon  similar  to  two  given  similar  polygons 
and  equivalent  to  their  sum.  (See  §  509.) 

Ex.  896.  Construct  a  polygon  similar  to  two  given  similar  polygons 
and  equivalent  to  their  difference. 

Ex.  897.  Construct  an  equilateral  triangle  equivalent  to  the  sum  of 
two  given  equilateral  triangles. 

Ex.  898.  Construct  an  equilateral  triangle  equivalent  to  the  differ- 
ence of  two  given  equilateral  triangles, 


BOOK   IV  237 

PROPOSITION  XIV.     PROBLEM 

513.   To   construct   a  polygon  similar  to    one  of  two 
given  polygons  and  equivalent  to  the  other. 


Given   polygons  P  and  Q,  with  s  a  side  of  P. 
To  construct   a  polygon  ~  P  and  =0=  Q. 

I.    Analysis 

1.  Imagine  the  problem   solved  and  let  R  be  the  required 
polygon  with  side  x  homol.  to  s,  a  side  of  P. 

2.  Then  P  :  R  =  s2  :  3? ;  i.e.  P  :  Q  =  s2  :  x~,  since  Q  =0=  R.     (1) 

3.  Now  to  avoid  comparing  polygons  which  are  not  similar, 
we  may  reduce  P  and  Q  to  =o  squares.     Let  the  sides  of  these 
squares  be  m  and  n,  respectively  ;  then  m2  =0=  P  and  n2  =0=  Q. 

4.  .-.  m2  :  w2  =  s2 :  or,  from  (1). 

5.  .•.  m  :  n  =  s  :  x. 

6.  That  is,  x  is  the  fourth  proportional  to  ra,  n,  and  s. 

II.  The  construction,  proof,  and  discussion  are  left  as  an 
exercise  for  the  student. 

514.  Historical  Note.  This  problem  was  first  solved  by  Pythagoras 
about  550  B.C. 

Ex.  899.  Construct  a  triangle  similar  to  a  given  triangle  and  equiva- 
lent to  a  given  parallelogram. 

Ex.  900.     Construct  a  square  equivalent  to  a  given  pentagon. 

Ex.  901.  Construct  a  triangle,  given  its  angles  and  its  area  (equal  to 
that  of  a  given  parallelogram).  HINT.  See  Prop.  XIV. 

Ex.  902.  Divide  a  triangle  into  two  equivalent  parts  by  a  line  drawn 
perpendicular  to  the  base.  HINT.  Draw  a  median  to  the  base,  then  apply 
Prop.  XIV. 


238 


PLANE   GEOMETRY 


Ex.  903.  Fig.  1  represents  maps  of  Utah  and  Colorado  drawn  to 
the  scale  indicated.  By 
carefully  measuring  the 
maps:  (1)  Calculate  the 
perimeter  of  each  state. 
(2)  Calculate  the  area  of 


each  state.  (3)  Check 
your  results  for  (2)  by 
comparing  with  the  areas 
given  for  these  states  in 
your  geography. 


U     T 


COLORADO 


SCALE  OF  MILES 
200          300          400 


500 


PENNSYLVANIA 

.  Pittsburg 

Harrisburg* 

Philadelphia  , 


FIG.  2. 


Ex.  904.     Fig.  2  repre-          <?    '? 

sents  a  map  of  Pennsylva-  pIG 

nia.    A  straight  line  from 
the  southwest  corner  to  the  northeast  corner  is 
300  miles  long. 

(1)  Determine  the  scale  to  which  the  map 
is  drawn. 

(2)  Calculate  the  distance  from  Pittsburg  to 
Harrisburg  ;  from  Harrisburg  to  Philadelphia  ; 
from  Philadelphia  to  Scran  ton  ;  from  Scranton 
to  Harrisburg. 

(3)  Calculate  approximately  the  area  of  the 
state  in  square  miles. 

Ex.  905.  Let  C  represent  a  can- 
dle ;  A  a  screen  1  foot  square  and 
1  yard  from  6Y;  -B  a  screen  2  feet 
square  and  2  yards  from  C ;  D  a 
screen  3  feet  square  and  3  yards 
from  (7.  If  screen  A  were  removed, 
the  quantity  of  light  it  received  would 
fall  on  B.  What  would  happen  if  B 
were  removed  ?  On  which  screen, 
then,  would  the  light  be  the  least 
intense?  From  the  figure,  determine 
the  law  of  intensity  of  light.  FIG.  3. 

Ex.  906.  Divide  a  triangle  into  two  equivalent  parts  by  a  line  drawn 
from  a  given  point  in  one  of  its  sides. 

HINT.  Let  M  be  the  given  point  in  AC  of  triangle  ABC ;  then 
\  AB  •  AC  =  AM  -  AX,  where  MX  is  the  required  line. 


BOOK   IV  239 

Ex.  907.  A  represents  a  station.  Cars  approach  the  station  on 
track  BA  and  leave  the  station  on  track  AC.  Construct  an  arc  of  a  circle 
DEj  with  given  radius  r,  connecting  the  two 
intersecting  car  lines,  and  so  that  each  car 
line  is  tangent  to  the  arc. 

This  same  principle  is  involved  in  designing 
a  building  between  two  streets  forming  at  their    £< 
point  of  intersection  a  small   acute  angle,  as 
the  Flatiron  Building  in  New  York  City. 

Ex.  908.  Find  the  area  of  a  rhombus  if 
its  diagonals  are  in  the  ratio  of  5  to  7  and  their 
sum  is  1C.  Fia-  4- 

MISCELLANEOUS  EXERCISES 

Ex.  909.  Show  that  if  a  and  b  are  two  sides  of  a  triangle,  the  area  is 
£  ab  when  the  included  angle  is  30°  or  150°;  |  ab\^  when  the  included 
angle  is  45°  or  135° ;  \  abV~3  when  the  included  angle  is  GO0  or  120°. 

Ex.  910.  The  sum  of  the  perpendiculars  from  any  point  within  a 
convex  polygon  upon  the  sides  is  constant. 

HINT.  Join  the  point  with  the  vertices  of  the  polygon,  and  consider 
the  sum  of  the  areas  of  the  triangles. 

Ex.  911.  The  sum  of  the  squares  on  the  segments  of  two  perpendicular 
chords  in  a  circle  is  equivalent  to  the  square  on  the  diameter. 

Ex.  912.  The  hypotenuse  of  a  right  triangle  is  20,  and  the  projection 
of  one  arm  upon  the  hypotenuse  is  4.  What  is  its  area? 

Ex.  913.  A  quadrilateral  is  equivalent  to  a  triangle  if  its  diagonals 
and  the  angle  included  between  them  are  respectively  equal  to  two  sides 
and  the  included  angle  of  the  triangle. 

Ex.  914.  Transform  a  given  triangle  into  another  triangle  containing 
two  given  angles. 

Ex.  915.  Prove  geometrically  the  algebraic  formula  (a  +  6)  (c  +  cT)  = 
ac  +  be  +  ad  +  bd. 

Ex.  916.  If  in  any  triangle  an  angle  is  equal  to  two  thirds  of  a  straight 
angle  (§  69),  then  the  square  on  the  side  opposite  is  equivalent  to  the  sum 
of  the  squares  on  the  other  two  sides  and  the  rectangle  contained  by  them. 

Ex.  917.  The  two  medians  .R/f  and  SH  of  the  triangle  RST  inter- 
sect at  P.  Prove  that  the  triangle  EPS  is  equivalent  to  the  quadrilateral 
HPKT. 

Ex.  918.  Find  the  area  of  a  triangle  if  two  of  its  sides  are  G  inches 
and  7  inches  and  the  included  angle  is  30°. 


240  PLANE   GEOMETRY 

Ex.  919.  By  two  different  methods  find  the  area  of  an  equilateral 
triangle  whose  side  is  10  inches. 

Ex.  920.  The  area  of  an  equilateral  triangle  is  36  \/3  ;  find  a  side  and 
an  altitude. 

Ex.  921.  By  using  the  formula  of  Prop.  IV,  Arg.  8,  derive  the 
formula  for  the  area  of  an  equilateral  triangle  whose  side  is  a. 

Ex.  922.  What  does  the  formula  for  T  in  Prop.  IV  become  if  angle 
C  is  a  right  angle  ? 

Ex.  923.  Given  an  equilateral  triangle  ABC,  inscribed  in  a  circle 
whose  center  is  0.  At  the  vertex  C  erect  a  perpendicular  to  BC  cutting 
the  circumference  at  D.  Draw  the  radii  OD  and  00.  Prove  that  the 
triangle  ODC  is  equilateral. 

Ex.  924.  Assuming  that  the  areas  of  two  triangles  which  have  an 
angle  of  the  one  equal  to  an  angle  of  the  other  are  to  each  other  as  the 
products  of  the  sides  including  the  equal  angles,  prove  that  the  bisector 
of  an  angle  of  a  triangle  divides  the  opposite  side  into  segments  propor- 
tional to  the  adjacent  sides. 

Ex.  925.  A  rhombus  and  a  square  have  equal  perimeters,  and  the 
altitude  of  the  rhombus  is  three  fourths  its  side ;  compare  the  areas  of  the 
two  figures. 

Ex.  926.  The  length  of  a  chord  is  10  feet,  and  the  greatest  perpen- 
dicular from  the  subtending  arc  to  the  chord  is  2  feet  1\  inches.  Find 
the  radius  of  the  circle. 

Ex.  927.  In  any  right  triangle  a  line  from  the  vertex  of  the  right  angle 
perpendicular  to  the  hypotenuse  divides  the  given  triangle  into  two  tri- 
angles similar  to  each  other  and  similar  to  the  given  triangle. 

Ex.  928.  The  bases  of  a  trapezoid  are  16  feet  and  10  feet,  respec- 
tively, and  each  of  the  non-parallel  sides  is  5  feet.  Find  the  area  of  the 
trapezoid.  Also  find  the  area  of  a  similar  trapezoid,  if  each  of  its  non- 
parallel  sides  is  3  feet. 

Ex.  929.  A  triangle  having  a  base  of  8  inches  is  cut  by  a  line  paral- 
lel to  the  base  and  6  inches  from  it.  If  the  base  of  the  smaller  triangle 
thus  formed  is  5  inches,  find  the  area  of  the  larger  triangle. 

Ex.  930.  If  the  ratio  of -similitude  of  two  similar  triangles  is  7  to  1, 
how  often  is  the  less  contained  in  the  greater?  HINT.  See  §§  418,  503. 

Ex.  931.     Construct  a  square  equivalent  to  one  third  of  a  given  square. 

Ex.  932.  If  the  side  of  one  equilateral  triangle  is  equal  to  the  alti- 
tude of  another,  what  is  the  ratio  of  their  areas  ? 

Ex.  933.    Divide  a  right  triangle  into  two  isosceles  triangles. 


BOOK  IV 


241 


EXERCISES   OF  GREATER  DIFFICULTY 

Ex.  934.     Construct  a  rectangle  equivalent  to  a  given  square  and  hav- 
ing the  sum  of  its  base  and  altitude  equal  to  a  given  line. 


R 


B  y  x 

/Is    '  -* 


Ex.  935.     Construct  a  rectangle  equivalent  to  a  given  triangle  and 
having  its  perimeter  equal  to  a  given  line. 

Ex.  936.     Construct  two  lines,  having  given  their  sum  and   their 
product. 

Ex.  937.     Construct  a  rectangle  equivalent  to  a  given  square  and  hav- 
ing the  difference  of  its  base  and  altitude  equal  to  a  given  line. 


L   *'          1 

E 

s 

I    A 

h 

• 
i 

//           i          T 

v   _  /..•:/>             /       -+-=+.  —  i  : 

8 

JJE^v       /          D  '1C"  
h%i^*'''                                 b 

C 

B~± 

&- 

Ex.  938.  Construct  a  rectangle  equivalent  to  a  given  rhombus,  the 
difference  of  the  base  and  altitude  of  the  rectangle  being  equal  to  a  given 
line. 

Ex.  939.  Construct  two  lines,  having  given  their  difference  and  their 
product. 

Ex.940.  Construct  a  triaijgle,  given  its  three  altitudes.  HINT.  Con- 
struct first  a  triangle  whose  sides  are  proportional  to  the  three  given  alti- 
tudes. 

Ex.  941.  Through  the  vertices  of  an  equilateral  triangle  draw  three 
lines  which  shall  form  an  equilateral  triangle  whose  side  is  equal  to  a 
given  line. 


242 


PLANE   GEOMETRY 


Ex.  942.  The  feet  of  the  perpendiculars  dropped  upon  the  sides  of  a 
triangle  from  any  point  in  the  circumference  of  the  circumscribed  circle 
are  collinear. 

rs 

OUTLINE  OF  PROOF.     The  circle  having  ^— -""Tt 

AP  as  diameter  will  pass  through  M  and  Q.  £*-^^— .  /  » 

.-.  Zl  -Z.V  and  Z2  =  Z2'.     Similarly 


Z  PKM  =  Z  PB  M. 


and 


.•.  QM  and  Jf/f  form  one  str.  line. 

Ex.  943.     Given  the  base,  the  angle  at  the  vertex,  and  the  sum  of  the 
other  two  sides  of  a  triangle;  construct  the  triangle. 


ANALYSIS.  Imagine  the  problem  solved  and  draw 
A  ABC.  Prolong  AB,  making  BE  =  BC,  since  the  line 
c  +  a  is  given.  Since  ZE—^ZABC,  A  AEC  can  be 
constructed. 

Ex.  944.  The  hypotenuse  of  a  right  triangle  is  given 
in  magnitude  and  position  ;  find  the  locus  of  the  center 
of  the  inscribed  circle. 


Ex.  945.     Prove  Prop.  VIII,  Book  IV,  by  using  the  following  figure, 
in  which  A'D'E'  is  placed  in  the  position  ADE. 


E' 


Ex.  946.  Prove  Prop.  VIII,  Book  IV,  using  two  triangles  such  that 
one  will  not  fall  wholly  within  the  other. 

Ex.  947.  If  two  triangles  have  an  angle  of  one  supplementary  to  an 
angle  of  the  other,  the  triangles  are  to  each  other  as  the  products  of  the 
sides  including  the  supplementary  angles.  (Prove  by  method  similar  to 
that  of  Ex.  945.) 


BOOK   IV 


243 


Ex.  948.     Given  base,    difference  of  sides,   and  difference   of  base 

angles  ;  construct  the  triangle. 

ANALYSIS.  In  the  accompanying  figure  suppose 
c  and  EA,  (/>  — a),  to  be  given.  A  consideration 
of  the  figure  will  show  that  Z2  =  Zl-f-ZA 
Add  Z  1  to  both  members  of  the  equation  ;  then 
Zl  +Z2  =  2Z1 +  ZA  ButZl  +  Z2  =  Z£.  B  c  ~~A 

.'.  Z7?  =  2Z1  +  ZA      .-.  Zl  =  \(Z.B  -  Z.A). 

The  A  BE  A  may  now  be  constructed.  The  rest  of  the  construction  is  left 
for  the  student. 

Ex.  949.     Given  base,  vertex  angle,  and  difference  of  sides,  construct 
the  triangle. 

ANALYSIS.     AE  =  AB-  EC.     Z.AEG=W°  B 

+  \  Z  B.     . •.  A  AEG  can  be  constructed. 

Ex.  950.  If  upon  the  sides  of  any  triangle 
equilateral  triangles  are  constructed,  the  lines 
joining  the  centers  of  the  equilateral  triangles 
form  an  equilateral  triangle. 

HINT.     Circumscribe  circles  about  the  three   A C 
equilateral  A.     Join  0,  the  common  point  of 

intersection  of  the  three  circles,  to  A,  B,  and  (7,  the  vertices  of  the  given  A. 
Prove  each  Z  at  0  the  supplement  of  the  Z  opposite  in  an  equilateral  A, 
and  also  the  supplement  of  the  Z  opposite  in  the  A  to  be  proved  equi- 
lateral. 


Ex.  951.     To  inscribe  a  square  in  a  semicircle. 
ANALYSIS.     Imagine  the  problem  solved  and  ABCD 

the  required  square.    Prove  OA  =  ——  .    Draw  EF  _L  EO 

2 

FE  :  EO  =  BA  :  AO. 


meeting  OB  prolonged  at  F. 
.'.  EF=2  OE. 


E   A  0  D 


Ex.  952.     To  inscribe  a  square  in  a  given 
triangle. 

OUT  LINK  OF  SOLUTION.  Imagine  the  prob- 
lem solved  and  ABCD  the  required  square-. 
Draw  >V/»'|I  ET  and  construct  square  KSFJL 
Draw  7?F,  thus  determining  point  (7.  The 
cons,  will  be  evident  from  the  figure.  To  prove  ABCD  a  square,  prove 
BC  =  CD.  BC:  SF  =  CD  :  FH.  But  SF  =  FH.  . :  BC  =  CD. 


R 


A    KD  T  H 


244 


PLANE   GEOMETRY 


E 


K 


Ex.  953.     In  a  given  square  construct  a  square,  having  a  given  side, 
so  that  its  vertices  shall  lie  in  the  sides  of  the  given  square. 

HINT.     Construct  a  rt.  A,  given  the  hypotenuse  and  sum  of  arms. 

Ex.  954.     Construct  a  triangle,  given  ma,  wc,  hb. 

ANALYSIS.     Imagine  the  problem  solved  and  R 

that  ABC  is  the  required  A.  If  BF  were  moved 
II  to  itself  till  it  contained  K,  the  rt.  A  AKT 
would  be  formed  and  KT  would  equal  \  BF. 

Then  A  AKT  can  be  made  the  basis  of  the       ^       ~~p       j>       £ 
required  construction. 

Ex.  955.     Construct  a  quadrilateral,  given  two  of  its  opposite  sides, 
its  two  diagonals,  and  the  angle  between  them. 

OUTLINE  OF  CONSTRUCTION.  Imagine  the 
problem  solved  and  that  ABCD  is  the 
required  quadrilateral,  s,  s',  d,  d',  and  /.DOC 
being  given.  By  II  motion  of  d  and  d'  the 
parallelogram  BKFD  may  be  obtained.  The 
required  construction  may  be  begun  by  draw- 
ing O  BKFD,  since  two  sides  and  the  in- 
cluded Z  are  known.  With  K  as  center  and  s 
as  radius,  describe  an  arc  ;  with  D  as  center  and 
arc  intersecting  the  first,  as  at  C, 
to  locate  A. 

Ex.  956.     Between  two  circles  draw  a  line  which  shall  be  parallel  to 
the  line  of  centers  and  equal  to  a  given  line  I. 


-^..  K 


'  as  radius,  describe  an 
Construct  EH  ABKG,  or  EH  DACF, 


Ex.  957.     Find  x  =  — ,  where  «,  ft,  c,  and  d  represent  given  lines. 
d1 


HINT. 


Here  x  =—  •  -•     Let— =y  and  construct 
d     d  d 


Then  construct  y. 


Ex.  958.     Transform  any  given  triangle  into  an  equilateral  triangle 
by  a  method  different  from  that  used  in  Ex.  865. 

ANALYSIS.     Call  the  base  of  the  given  triangle  ft  and  its  altitude  a. 
Let  x  =  the  side  of  the  required  equilateral  triangle. 

Then   —  V3  =  -6-^.     .'.  \  b  :  x  =  3 
4  2 


BOOK   Y 

REGULAR   POLYGONS.     MEASUREMENT   OF   THE 
CIRCLE 

515.  Def.     A  regular  polygon  is  one  which  is  both  equilateral 
and  equiangular. 

Ex.  959.     Draw  an  equilateral  triangle.     Is  it  a  regular  polygon  ? 

Ex.  960.  Draw  a  quadrilateral  that  is  equilateral  but  not  equiangular ; 
equiangular  but  not  equilateral ;  neither  equilateral  nor  equiangular ;  both 
equilateral  and  equiangular.  Which  of  these  quadrilaterals  is  a  regular 
polygon  ? 

Ex.  961.  Find  the  number  of  degrees  in  an  angle  of  a  regular  do- 
decagon. 

516.  Historical  Note.    The  following  theorem  presupposes  the  pos- 
sibility of  dividing  the  circumference  into  a  number  of  equal  arcs.     The 
actual  division  cannot  be  obtained  by  the  methods  of  elementary  geom- 
etry, except  in  certain  special  cases  which  will  be  discussed  later. 

As  early  as  Euclid's  time  it  was  known  that  the  angular  magnitude 
about  a  point  (and  hence  a  circumference)  could  be  divided  into  2n, 
2n  •  3,  2"  •  5,  2n  •  15  equal  angles.  In  1706  it  was  discovered  by  Gauss, 
then  nineteen  years  of  age,  that  a  regular  polygon  of  17  sides  can  be  con- 
structed by  means  of  ruler  and  compasses,  and  that  in  general  it  is  possi- 
ble to  construct  all  polygons  having  (2"  +  1)  sides,  n  being  an  integer 
and  (2"+  1)  a  prime  number.  The  first  four  numbers  satisfying  this 
condition  are  3,  5,  17,  257.  Gauss  proved  also  that  polygons  having  a 
number  of  sides  equal  to  the  product  of  two  or  more  different  numbers 
of  this  series  can  be  constructed. 

Gauss  proved,  moreover,  that  only  a  limited  class  of  regular  polygons  are 
constructible  by  elementary  geometry.  For  a  note  on  the  life  of  Gauss, 
see  §  520. 

245 


246 


PLANE   GEOMETRY 


PROPOSITION  I.     THEOREM 

517.  If  the  circumference  of  a  circle  is  divided  into 
any  number  of  equal  arcs:  (a)  the  chords  joining  the 
points  of  division  form  a  regular  polygon  inscribed  in 
the  circle ;  (b)  tangents  drawn  at  the  points  of  division 
form  a  regular  polygon  circumscribed  about  the  circle. 


Given  circumference  AGE  divided  into  equal  arcs  AB,  BC,  CD, 
etc.,  and  let  chords  AB,  BC,  CD,  etc.,  join  the  several  points  of 
division,  and  let  the  tangents  GH,  HK,  KL,  etc.,  touch  the  circum- 
ference at  the  several  points  of  division. 

To  prove   A  BCD  •  -  •  and  GHKL  •  •  •  regular  polygons. 


1. 
2. 

3. 
4. 

5. 
6. 

7. 

8. 
9. 
10. 


ARGUMENT 

AB  =  BC=  CD  =  -  •  -  . 

.-.  arc  CEA  =  arc  DFB  =  arc  EAC  = 


Also  AB  =  BC=  CD  =  •••. 

.-.  ABCD  •  •  •  is  a  regular  polygon. 

Again,    ZBAG  =  Z.GBA  =   ZCBH 

Z.HCB  =  •  -  .  . 
And  AB  =  BC=  CD=-  ••. 

.-.  AAGB  = 


And  AG  =  GB  =  BH= 


REASONS 

1.  By  hyp. 

2.  §54,  7  a. 

3.  §  362,  a. 

4.  §  298. 

5.  §  515. 

6.  §  362,  a. 


7.  Arg.  4. 

8.  §  105. 

9.  §  110. 
10.  §  110. 


BOOK  V 


247 


ARGUMENT  REASONS 

11.  .'.  GH=HK=KL=-..  11.    §54,  7  a. 

12.  .-.  GHKL-  •  •  is  a  regular  polygon.  Q.E.D.      12.    §515. 

518.  Questions.     If,  in  the  figure  of  §  517,  the  circumference  is  di- 
vided into  six  equal  parts,  how  many  arcs,  each  equal  to  arc  AB,  will  arc 
CEA  contain  ?   arc  DFB  ?     How  many  will  each  contain  if  the  circum- 
ference is  divided  into  n  equal  parts  ?     In  step  10,  why  does  ACr  =  GB  ? 

519.  Cor.     //  the  vertices  of  a  regular  inscribed  poly- 
gon are  joined  to  the  jnid-points  of  the  arcs  subtended  by 
the  sides  of  the  polygon,  tlie  joining  lines  will  form  a 
regular  inscribed  polygon  of  double  tlxe  number  of  sides. 

Ex.  962.     An  equilateral  polygon  inscribed  in  a  circle  is  regular. 
Ex.  963.     An  equiangular  polygon  circumscribed  about  a  circle  is 
regular. 

520.  Historical  Note.     Karl  Friedrich  Gauss  (1777-1855)  was  born 
at  Brunswick,  Germany.    Although  he  was  the  son  of  a  bricklayer,  he  was 
enabled  to  receive  a  liberal  ed- 
ucation, owing  to  the  recogni- 
tion of  his  unusual  talents  by  a 

nobleman.  He  was  sent  to  the 
Caroline  College  but,  at  the  age 
of  fifteen,  it  was  admitted  both 
by  professors  and  pupils  that 
Gauss  already  knew  all  that 
they  could  teach  him.  He 
became  a  student  in  the  Uni- 
versity of  Gottingen  and  while 
there  did  some  important  work 
on  the  theory  of  numbers. 

On  his  return  to  Brunswick, 
he  lived  humbly  as  a  private 
tutor,  until  1807,  when  he 
was  appointed  professor  of 

astronomy  and  director  of  the  observatory  at  Gottingen.  While  there  he 
did  important  work  in  physics  as  well  as  in  astronomy.  He  also  invented 
the  telegraph  independently  of  S.  F.  B.  Morse. 

His  lectures  were  unusually  clear,  and  he  is  said  to  have  given  in  them 
the  analytic  steps  by  which  he  developed  his  proofs  ;  while  in  his  writings 
there  is  no  hint  of  the  processes  by  which  he  discovered  his  results. 


GAUSS 


248 


PLANE    GEOMETRY 


PROPOSITION  II.     PROBLEM 
521.   To  inscribe  a  square  in  a  given  circle. 


Given    circle  0. 

To  inscribe   a  square  in  circle  0. 

I.    The  construction  is  left  as  an  exercise  for  the  student. 

II.    Proof 
ARGUMENT  REASONS 

1.  AB  _L  CD.  1.    By  cons. 

2.  .-.  Z  AOC=  90°.  2.    §  71. 

3.  .-.  AC  =  90°,   i.e.  one  fourth  of  the  cir-     3.    §  358. 

cu  inference. 

4.  .•.  the  circumference  is  divided  into  four     4.    Arg.  3. 

equal  parts. 

5.  .-.  polygon  ACBD,  formed  by  joining  the      5.    §  517,  a. 

points  of  division,  is  a  square.    Q.E.D. 

III.    The  discussion  is  left  as  an  exercise  for  the  student. 

522.  Cor.  The  side  of  a  square  inscribed  in  a  circle  is 
equal  to  the  radius  multiplied  by  ~\T2 ,'  the  side  of  a  square 
circumscribed  about  a  circle  is  equal  to  twice  the  radius. 


Ex.  964.     Inscribe  a  regular  octagon  in  a  circle. 

Ex.  965.     Inscribe  in  a  circle  a  regular  polygon  of  sixteen  sides. 

Ex.  966.     Circumscribe  a  square  about  a  circle. 


BOOK   V 


249 


Ex.  967.     Circumscribe  a  regular  octagon  about  a  circle. 
Ex.  968.     On  a  given  line  as  one  side,  construct  a  square. 
Ex.  969.     On  a  given  line  as  one  side,  construct  a  regular  octagon. 
Ex.  970.     If  a  is  the  side  of  a  regular  octagon  inscribed  in  a  circle 
whose  radius  is  S,  then  a  =  E  V 2  —  \/2. 

PROPOSITION  III.     PROBLEM 
523.    To  inscribe  a  regular  hexagon  in  a  given  circle. 


Given   circle  0. 

To  inscribe   in  circle  O  a  regular  hexagon. 

I.   The  construction  is  left  as  an  exercise  for  the  student. 
HINT.     AB  =  radius  OA: 

II.    Proof 

ARGUMENT 

1.  Draw  OB. 

2.  Then  A  AB  0  is  equilateral. 
.-.  Zo=60°. 

.-.  AB  =  60°,  i.e.  one  sixth  of  the  cir- 
cumference. 

.*.  the  circumference  may  be/  divided 
into  .six  equal  parts. 

.-.  polygon  ABCDEF,  formed  by  joining 
the  points  of  division,  is  a  regular 
inscribed  hexagon.  Q.E.D. 

III.     The  discussion  is  left  as  an  exercise  for  the  student. 


6. 


REASONS 

1.  §  54,  15. 

2.  By  cons. 

3.  §213. 

4.  §358. 

5.  Arg.  4. 

6.  §  517,  a. 


250 


PLANE   GEOMETRY 


524.  Cor.  I.    A  regular  inscribed  triangle  is  formed  by 
joining  the  alternate  vertices  of  a 

regular  inscribed  liexagon. 

525.  Cor.  II.     A  side  of  a  regu- 
lar inscribed  triangle  is  equal  to 
the  radius  of  the  circle  multi- 
plied by  V 3. 

HINT.    A  ABD  is  a  right  triangle  whose 
hypotenuse  is  2  R  and  one  side  B. 


Ex.  971.     Inscribe  a  regular  dodecagon  in  a  circle. 

Ex.  972.  Divide  a  given  circle  into  two  segments  such  that  any  angle 
inscribed  in  one  segment  is  five  times  an  angle  inscribed  in  the  other. 

Ex.  973.     Circumscribe  an  equilateral  triangle  about  a  circle. 

Ex.  974.     Circumscribe  a  regular  hexagon  about  a  circle. 

Ex.  975.     On  a  given  line  as  one  side,  construct  a  regular  hexagon. 

Ex.  976.     On  a  given  line  as  one  side,  construct  a  regular  dodecagon. 

Ex.  977.  If  a  is  the  side  of  a  regular  dodecagon  inscribed  in  a  circle 
whose  radius  is  H,  then  a  =  R\2  —  V3. 

PROPOSITION  IV.     PROBLEM 
526.    To  inscribe  a  regular  decagon  in  a  circle. 


^E 


X 

FIG.  1. 

Given    circle  0. 

To  inscribe   in  circle  0  a  regular  decagon. 


FIG.  2. 


BOOK  V 


251 


I.    Construction 

1.  Divide  a  radius  R,  of  circle  0,  in  extreme  and  mean  ratio. 
§  465.     (See  Fig.  2.) 

2.  In  circle  O  draw  a  chord  AB,  equal  to  s,  the  greater  seg- 
ment of  R. 

3.  AB  is  a  side  of  the  required  decagon. 


1. 
2. 
3. 
4. 
5. 
6. 

7. 
8. 

9. 
10. 
11. 
12. 

13. 

14. 

15. 

16. 

17. 
18. 

19. 
20. 


II.    Proof 
ARGUMENT 

Draw  radius  OA. 

On  OA  lay  off  OM  =  s. 

Draw  OB  and  BM. 

Then  OA  :  OM  =  OM  :  MA. 

.'.         OA  :  AB  =  AB  :  MA. 

Also  in  A  OAB  and  MAB,  Z  A  =  Z  ^4. 

.'.A  OAB  ~  A  MAB. 

v  A  OAB  is  isosceles,  A  MAB  is  isosceles, 

and  -W  =  AB. 
But    ^B  =  s  =  OM. 
.-.        BM=  OM. 

.-.  A  50J/  is  isosceles  and  Z.MBO  =  /-O. 
But  Z  ,47?  J/  =  Z  o. 

GT  /-AKO  =2ZO. 


In    A  ABO,  Z.ABO  -f  /.MAB 

180°. 
/.  2  Z  0  +  2  Z  0  +  Z  0,  or  5  Z  0,  = 

180°. 

.-.  Zo  =  36°. 
.-.  AB  =  36°,  i.e.  one  tenth  of  the  cir- 

cumference. 
/.  the  circumference  may  be  divided 

into  ten  equal  parts. 
/.  polygon  ABCD  •  •  •  ,  formed  by  join- 

ing the  points  of  division,  is  a  regular 

inscribed  decagon.  Q.E.D. 


11. 
12. 


14. 
15. 


REASONS 
§  54,  15. 
§54,14. 
§  54,  15. 
By  cons. 
§309. 
By  iden. 
§428. 
§94. 


9.    By  cons. 
10.    §54,1. 
§111. 
§424,1. 


13.    §  54,  2. 
111. 
204. 


16.  §309. 

17.  §54,  8  a. 

18.  §358. 

19.  Arg.  18. 

20.  §  517,  a. 


252  PLANE   GEOMETRY 

III.    The  discussion  is  left  as  an  exercise  for  the  student. 

527.   Cor.  I.     A  regular  pentagon  is  formed  by  joining 
the  alternate  vertices  of  a  regular  inscribed  decagon. 


Ex.  978.     Construct  a  regular  inscribed  polygon  of  20  sides. 

Ex.  979.     The  diagonals  of  a  regular  inscribed  pentagon  are  equal. 

Ex.  980.     Construct  an  angle  of  36°  ;   of  72°. 

Ex.  981.     Divide  a  right  angle  into  five  equal  parts. 

Ex.  982.  The  eight  diagonals  of  a  regular  decagon  drawn  from  any 
vertex  divide  the  angle  at  that  vertex  into  eight  equal  angles. 

Ex.  983.     Circumscribe  a  regular  pentagon  about  a  circle. 

Ex.  984.     Circumscribe  a  regular  decagon  about  a  circle. 

Ex.  985.     On  a  given  line  as  one  side,  construct  a  regular  pentagon. 

Ex.  986.     On  a  given  line  as  one  side,  construct  a  regular  decagon. 

Ex.  987.  The  side  of  a  regular  inscribed  decagon  is  equal  to 
|  B  (V5  —  1),  where  E  is  the  radius  of  the  circle. 

HINT.     By  cons.,  H  :  s  —  s  :  K  —  s.    Solve  this  proportion  for  s. 

PROPOSITION  V.     PROBLEM 
528.    To  inscribe  a  regular  pentedecagon  in  a  circle. 


Given    circle  O. 

To  inscribe  in  circle  O  a  regular  pentedecagon. 

I.    Construction 

1.  Prom  A,  any  point  in  the  circumference,  lay  off  chord  AK 
equal  to  a  side  of  a  regular  inscribed  hexagon.     §  523. 


BOOK  V  253 

2.  Also  lay  off  chord  AF  equal  to  a  side  of  a  regular  inscribed 
decagon.     §  526. 

3.  Draw  chord  FE. 

4.  FE  is  a  side  of  the  required  pentedecagori. 


II.    Proof 

ARGUMENT 

1.  Arc  AE  =  ^  of  the  circumference. 

2.  Arc  AF  =  TL  of  the  circumference. 

3.  .'.  arc  FE  =  i  —  TV>  «'•€.  y1^  of  the  cir- 

cumference. 

4.  .-.  the    circumference   may  be   divided 

into  fifteen  equal  parts. 

5.  .•.  the  polygon  formed   by  joining  the 

points  of  division  will  be  a  regular 
inscribed  pentedecagon.  Q.E.D. 


REASONS 

1.  By  cons. 

2.  By  cons. 

3.  §  54,  3. 

4.  Arg.3. 


5.    §  517,  a. 


III.   The  discussion  is  left  as  an  exercise  for  the  student. 

529.    Note.     It  has  now  been  shown  that  a  circumference  can  be 
divided  into  the  number  of  equal  parts  indicated  below  : 

2,  4,     8,     16,.-.  2»n 

3,  6,  12,     24,  •••    3  x  2n  I    [w  being  any  positive 
5,  10,  20,     40,  ...    5  x  2'1  |       integer]. 

15,  30,  60,  120,  ...  15  x  2»J 


Ex.  988.     Construct  an  angle  of  24°. 

Ex.  989.     Circumscribe  a  regular  pentedecagon  about  a  given  circle. 

Ex.  990.  On  a  given  line  as  one  side,  construct  a  regular  pente- 
decagon. . 

Ex.  991.  Assuming  that  it  is  possible  to  inscribe  in  a  circle  a 
regular  polygon  of  17  sides,  show  how  it  is  possible  to  inscribe  a  regular 
polygon  of  51  sides. 

Ex.  992.  If  a  regular  polygon  is  inscribed  in  a  circle,  the  tangents 
drawn  at  the  mid-points  of  the  arcs  subtended  by  the  sides  of  the  in- 
scribed polygon  form  a  circumscribed  regular  polygon  whose  sides  are 
parallel  to  the  sides  of  the  inscribed  polygon,  and  whose  vertices  lie  on 
the  prolongations  of  the  radii  drawn  to  the  vertices  of  the  inscribed 
polygon. 


254 


PLANE   GEOMETRY 


PROPOSITION  VI.     THEOREM 

530.   A  circle  may  be  circumscribed  about  any  regular 
polygon ;  and  a  circle  may  also  be  inscribed  in  it. 


At- 


Given   regular  polygon  ABCD 

To  prove :   (a)  that  a  circle  may  be  circumscribed  about  it ; 
(6)  that  a  circle  may  be  inscribed  in  it. 

(a)  ARGUMENT 

1.  Pass  a  circumference  through  points 

A,  B,  and  C. 

2.  Connect  0,  the  center   of   the   circle, 

with  all  the  vertices  of  the  polygon. 

3.  Then      OB '=  OC. 

4.  .-.  Z1  =  Z2. 

5.  I  Jut  Z  ABC  =  Z  BCD. 

6.  .-.  Z3  =  Z4.~ 

7.  Also       AB  =  CD. 

8.  .'.  AABO=  A  OCD. 

9.  .-.OA  =  ^)D    and    circumference   ABC 

passes  through  D. 

10.  In  like  manner  it  may  be  proved  that 
circumference  ABC  passes  through 
each  of  the  vertices  of  the  regular 
polygon ;  the  circle  will  then  be 
circumscribed  about  the  polygon. 

Q.E.D. 


KEASONS 

1. 

§  324. 

2.. 

§  54,  15. 

3. 

§  279,  a. 

4. 

§  111. 

5. 

§  515. 

6. 

§  54,  3. 

7. 

§  515. 

8. 

§  107. 

9. 

§  no. 

10. 

By      steps 

similar   to 

1-9. 

BOOK   V 


255 


b\  ARGUMENT 

1.  Again   AB,  BC,  CD,  etc.,  the  sides  of 

the  given  polygon,  are  chords  of 
the  circumscribed  circle. 

2.  Hence  J§  from  the  center  of  the  circle 

to  the§g_  chords  are  equal. 

3.  .-.  with  0  as  center,  and  with  a  radius 

equal  to  one  of  these  Js,  as  OH,  a 
circle  may  be  described  to  which  all 
the  sides  of  the  polygon  will  be 
tangent^ 

4.  .-.  this  circle  will  be  inscribed  in  the 

polygon.  Q.E.D. 


REASONS 

1.  §  281. 

2.  §  307. 

3.  §  314. 


4.    §  317. 


531.  Def.     The  center  of  a  regular  polygon  is  the  common 
center  of  the  circumscribed  and  inscribed  circles ;  as  0,  Prop.  VI. 

532.  Def.     The  radius  of  a  regular  polygon  is  the  radius  of 
the  circumscribed  circle,  as  OA. 

533.  Def.     The  apothem  of  a  regular  polygon  is  the  radius 
of  the  inscribed  circle,  as  OH. 

534.  Def.      In  a  regular  polygon  the  angle  at  the  center  is 
the  angle  between  radii  of  the  polygon  drawn  to  the  extremities 
of  any  side,  as  Z  AOF. 

535.  Cor.  I.  ffhe  angle  at  the  center  is  equal  to  four 
right   angles  Divided  by  the  number   of  sides   of  the 
polygon. 

536.  Cor.  II.     An  angle  of  a  regular  polygon  is  the 
supplement  of  the  angle  at  the  center. 


Ex.  993.  Find  the  number  of  degrees  in  the  angle  at  the  center  of  a 
regular  octagon.  Find  the  number  of  degrees  in  an  angle  of  the  octagon. 

Ex.  994.  If  the  circle  circumscribed  about  a  triangle  and  the  circle 
inscribed  in  it  are  concentric,  the  triangle  is  equilateral. 

Ex.  995.  How  many  sides  has  a  regular  polygon  whose  angle  at  the 
center  is  30°  ? 


256 


PLANE   GEOMETRY 


PROPOSITION   VII.     THEOREM 

537.  Regular  polygons  of  the  same  number  of  sides  are 
similar. 

C 


Given  two  regular  polygons,  ABCDE  and  A'B'C'D'E',  of  the 
same  number  of  sides. 

To  prove    polygon  ABCDE  ~  polygon  A'B'C'D'E'. 


ARGUMENT 

1.  Let  n  represent  the  number  of  sides  of 

each   polygon;    then   each   angle   of 
each  polygon  equals 

(rt-2)2rt.  A 
n 

2.  .'.    the    polygons    are    mutually    equi- 

angular. 

3.  AB  =  BC  =  CD  =  •  -  •  . 

4.  A'B'  =  B'C'  =  C'D'  =  •  •  •  . 

AB  __  BC  __  CD_  _ 

A\  7>f       pf  f,i       ft1  /t1 
A  11         x>  O          C  1) 


5. 


6.    .-.  polygon  ABCDE  ~  polygon  A'B'C'D'E'. 

Q.E.D. 


REASONS 


1.    §217. 


2.  Arg.  1. 

3.  §515. 

4.  §  515. 

5.  §  54,  8  a. 

6.  §419. 


Ex.  996.  Two  homologous  sides  of  two  regular  pentagons  are  3 
inches  and  5  inches,  respectively  ;  what  is  the  ratio  of  their  perimeters  ? 
of  their  areas  ? 

Ex.  997.  The  perimeters  of  two  regular  hexagons  are  30  inches  and 
72  inches,  respectively  ;  what  is  the  ratio  of  their  areas  ? 


BOOK  V 


257 


PROPOSITION  VIII.     THEOREM 

538.  The  perimeters  of  two  regular  polygons  of  the 
same  number  of  sides  are  to  each  other  as  their  radii  or 
as  their  apothems.  * 


E 


F'    E' 


Given  regular  polygons  ABODE  and  A'B'C'D'E'  having  the 
same  number  of  sides.  Let  OA  and  O'A'  be  radii,  OF  and  O'F' 
apothems,  P  and  p'  perimeters,  of  the  two  polygons,  respectively. 

P         OA         OF 
To  prove     —  —  — — -  =  — — -. 

p'     O'A'     O'F' 
The  proof  is  left  as  an  exercise  for  the  student. 


HINT.     From  §§  537  and  441,  —  = 

_£ 


(§535  and  §428).    Then 


AE 

A'E< 


OA 
O'A1 


'E± 

.  Of 
O'F' 


Prove    &AOE~&A'0'E' 
(§  435). 


539.  Cor.  The  areas  of  two  regular  polygons  of  tlie  same 
number  of  sides  are  to  each  other  as  the  squares  of  their 
radii  or  as  the  squares  of  their  apothems. 


Ex.  998.  Two  regular  hexagons  are  inscribed  in  circles  whose  radii 
are  7  inches  and  8  inches,  respectively.  Compare  their  perimeters.  Com- 
pare their  areas. 

Ex.  999.  The  lines  joining  the  mid-points  of  the  radii  of  a  regular 
pentagon  form  a  regular  pentagon  whose  area  is  one  fourth  that  of  the 
first  pentagon. 


258  PLANE   GEOMETRY 

MEASUREMENT  OF  THE  CIRCUMFERENCE  AND  OF 
THE   CIRCLE 

540.  The   measure   of    a   straight   line,  i.e.   its   length,  is 
obtained  by  laying  off  upon  it  a  straight  line  taken  as  a  stand- 
ard or  unit  (§  335). 

Since  a  straight  line  cannot  be  made  to  coincide  with  a 
curve,  it  is  obvious  that  some  other  system  of  measurement 
must  be  adopted  for  the  circumference.  The  following 
theorems  will  develop  the  principles  upon  which  such  measure- 
ment is  based. 

PROPOSITION  IX.     THEOREM 

541.  I.    The  perimeter  and  area  of  a  regular  polygon 
inscribed  in  a  circle  are  less,  respectively,  than  the  perime- 
ter and  area  of  the  regular  inscribed  polygon  of  twice  as 
many  sides. 

II.  The  perimeter  and  area  of  a  regular  polygon  cir- 
cumscribed about  a  circle  are  greater,  respectively,  than 
the  perimeter  and  area  of  the  regular  circumscribed  poly- 
gon of  twice  as  many  sides. 


The  proof  is  left  as  an  exercise  for  the  student. 

HINT.     The  sum  of  two  sides  of  a  triangle  is  greater  than  the  third  side. 


Ex.  1000.     A  square  and  a  regular  octagon  are  inscribed  in  a  circle 
whose  radius  is  10  inches  ;  find  : 

(«)  The  difference  between  their  perimeters. 
(6)  The  difference  between  their  areas. 


BOOK  V 


259 


ARCHIMEDES 


542.  Historical  Note.  Archimedes  (285?-212  B.C.)  found  the 
circumference  and  area  of  the  circle  by  a  method  similar  to  that  given  in 
this  text.  He  was  born  in  Syracuse,  Sicily,  but  studied  in  Egypt  at  the 
University  of  Alexandria. 

Although,  like  Plato,  he  re- 
garded practical  applications  of 
mathematics  as  of  minor  im- 
portance, yet,  on  his  return  to 
Sicily  he  is  said  to  have  won 
the  admiration  of  King  Hiero 
by  applying  his  extraordinary 
mechanical  genius  to  the  con- 
struction of  war-engines  with 
which  great  havoc  was  wrought 
on  the  Roman  army.  By 
means  of  large  lenses  arid  mir- 
rors he  is  said  to  have  focused 
the  sun's  rays  and  set  the 
Roman  ships  on  fire.  Although 
this  story  may  be  untrue,  never- 
theless such  a  feat  would  be  by  no  means  impossible. 

Archimedes  invented  the  Archimedes  screw,  which  was  used  in  Egypt 
to  drain  the  fields  after  the  inundations  of  the  Nile.  A  ship  which  was 
so  large  that  Hiero  could  not  get  it  launched  was  moved  by  a  system  of 
cogwheels  devised  by  Archimedes,  who  remarked  in  this  connection  that 
had  he  but  a  fixed  fulcrum,  he  could  move  the  world  itself. 

The  work  most  prized  by  Archimedes  himself,  however,  and  that  which 
gives  him  rank  among  the  greatest  mathematicians  of  all  time,  is  his  in- 
vestigation of  the  mechanics  of  solids  and  fluids,  his  measurement  of  the 
circumference  and  area  of  the  circle,  and  his  work  in  solid  geometry 

Archimedes  was  killed  when  Syracuse  was  captured  by  the  Romans. 
The  story  is  told  'that  he  was  drawing  diagrams  in  the  sand,  as  was  the 
custom  in  those  days,  when  the  Roman  soldiers  came  upon  him.  He 
begged  them  not  to  destroy  his  circles,  but  they,  not  knowing  who  he  was, 
and  thinking  that  he  presumed  to  command  them,  killed  him  with  their 
spears.  The  Romans,  directed  by  Marcellus,  who  admired  his  genius  and 
had  given  orders  that  he  should  be  spared,  erected  a  monument  to  his 
memory,  on  which  were  engraved  a  sphere  inscribed  in  a  cylinder. 

The  story  of  the  re-discovery  of  this  tomb  in  75  B.C.  is  delightfully  told 
by  Cicero,  who  found  it  covered  with  rubbish,  when  visiting  Syracuse. 

Archimedes  is  regarded  as  the  greatest  mathematician  the  world  has 
known,  with  the  sole  exception  of  Newton. 


260 


PLANE   GEOMETRY 


PROPOSITION  X.     THEOREM 

543.  By  repeatedly  doubling  the  number  of  sides  of 
a  regular  polygon  inscribed  in  a  circle,  and  making  the 
polygons  always  regular: 

I.  The  apothem  can  be  made  to  differ  from  the  ra- 
dius by  less  than  any  assigned  value. 

II.  The  square  of  the  apothem  can  be  made  to  differ 
from  the  square  of  the  radius  by  less  than  any  assigned 
value. 


Given   AB  the  side,  OC  the  apothem,  and  OB  the  radius  of  a 
regular  polygon  inscribed  in  circle  AMB. 

To  prove   that  by  repeatedly  doubling  the  number  of  sides 
of  the  polygon : 

I.    OB  —  OC  can  be  made  less  than  any  assigned  value. 

II.    OB*  —  OC2  can  be  made  less  than  any  assigned  value. 


I. 


ARGUMENT 


2. 


By  repeatedly  doubling  the  number  of 
sides  of  the  inscribed  polygon  and 
making  the  polygons  always  regular, 
AB}  subtended  by  one  side  AB  of  the 
polygon,  can  be  made  less  than  any 
previously  assigned  arc,  however  small. 

.•.  chord  AB  can  be  made  less  than  any 
previously  assigned  line  segment,  how- 
ever small. 


1. 


REASONS 
§519. 


§  301. 


BOOK  V 


261 


4. 


ARGUMENT 
•.  CB,  which  is  %AB,  can  be  made  less 

than  any  previously  assigned  value, 

however  small. 
But  OB  —  OO  <  CB. 
.-.  OB  —  OC,  being  always  less  than  CB, 

can  be  made  less  than  any  previously 

assigned  value,  however  small. 

Q.E.D. 
II. 

Again,  "off  —  oc2  —  CB2. 

But  CB  can  be  made  less  than  any  pre- 
viously assigned  value,  however  small. 

.•.  CB  can  be  made  less  than  any  previ- 
ously assigned  value,  however  small. 
OB2  —  OC2,  being  always  equal  to  CB2, 


3. 


REASONS 
544. 


4.  §  168. 

5.  §  59,  10. 


§  447. 
I,  Arg.  3. 


3.    §  545. 


4.    §  309. 


can  be  made  less  than  any  previously 
assigned  value,  however  small. 

Q.E.D. 

544.  If  a  variable  can  be  made  less  than  any  assigned  value, 
the  quotient  of  the  variable  by  any  constant,  except  zero,  can  be 
made  less  than  any  assigned  value. 

545.  If  a  variable  can  be  made  less  than  any  assigned  value, 
the  square  of  that  variable  can  be  made  less  than  any  assigned 
value.      (For  proofs  of  these  theorems  see  Appendix,  §§  586  and  589.) 


Ex.  1001.  Construct  the  following  designs:  (1)  on  the  blackboard, 
making  each  line  12  times  as  long  as  in  the  figure  ;  (2)  on  paper,  making 
each  line  4  times  as  long : 


262 


PLANE   GEOMETRY 
PROPOSITION  XI.     THEOREM 


546.  By  repeatedly  doubling  the  number  of  sides  of 
regular  circumscribed  and  inscribed  polygons  of  the 
same  number  of  sides,  and  making  the  polygons  always 
regular  : 

I.   Their  perimeters  approach  a  common  limit. 

II.   Their  areas  approach  a  common  limit. 


Given  P  and  p  the  perimeters,  R  and  r  the  apothems,  and  K 
and  k  the  areas  respectively  of  regular  circumscribed  and 
inscribed  polygons  of  the  same  number  of  sides. 

To  prove   that  by  repeatedly  doubling  the  number  of  sides 
of  the  polygons,  and  making  the  polygons  always  regular : 
I.    P  and  p  approach  a  common  limit. 

II.    K  and  k  approach  a  common  limit. 


I.  ARGUMENT 

1.    Since  the  two  regular  polygons  have  the 


2. 


3. 


P       R 

same  number  of  sides,  —  =  — . 
p      r 

P—p_R—r 
P  R 

R—r 


P—p  = 


K 


4.  But  by  repeatedly  doubling  the  number 
of  sides  of  the  polygons,  and  making 
them  always  regular,  R  —  r  can  be 


KEASONS 
1.   §  538. 


2.  §  399. 

3.  §  54,  7  a. 

4.  §  543,  I. 


BOOK   V 


263 


5. 


ARGUMENT 

made   less    than   any    previously   as- 
signed value,  however  small. 
R  —  r 


R 


can  be  made  less  than  any  pre- 


viously assigned  value,  however  small. 

T5    n* 

P can  be  made  less  than  any 

R 

previously   assigned   value,    however 
small,  P  being  a  decreasing  variable. 

P  —p,  being  always  equal  to  P  — -  — , 

can  be  made  less  than  any  previously 
assigned  value,  however  small. 
P  and  p  approach  a  common  limit. 

Q.E.D. 


REASONS 


5.    §  544. 


6.    §  547. 


7.    §  309. 


8.    §  548. 


II.     The  proof  of  II  is  left  as  an  exercise  for  the  student. 
HINT.     Since  the  two  regular  polygons  have  the  same  number  of  sides, 

K=*P.  (§  539).    The  rest  of  the  proof  is  similar  to  steps  2-8,  §  546,  I. 
k       r'2 

547.  If  a  variable  can  be  made  less  than  any  assigned  value, 
the  product  of  that  variable  and  a  decreasing  value  may  be  made 
less  than  any  assigned  value. 

548.  If  two   related   variables  are  such   that  one   is  always 
greater  than  the  other,  and  if  the  greater  continually  decreases 
while  the  less  continually  increases,  so  that  the  difference  between 
the  two  may  be  made  as  small  as  ive  please,  then  the  two  variables 
have  a  common  limit  tuhich  lies  between  them. 

(For  proofs  of  these  theorems  see  Appendix,  §§  587  and  594.) 

549.  Note.     The  above  proof  is  limited  to  regular  polygons,  but  it 
can  be  shown  that  the  limit  of  the  perimeter  of  any  inscribed  (or  circum- 
scribed) polygon  is  the  same  by  whatever  method  the  number  of  its  sides 
is  successively  increased,  provided  that  each  side  approaches  zero  as  a  limit. 


264 


PLANE   GEOMETRY 


550.  Def.     The  length  of   a  circumference  is   the   common 
limit  which  the  successive  perimeters  of  inscribed  and  circum- 
scribed regular  polygons  (of  3,  4,  5,  etc.,  sides)  approach  as  the 
number  of  sides  is  successively  increased  and  each  side  ap- 
proaches zero  as  a  limit. 

The   term   "  circumference "   is   frequently   used   for   "  the 
length  of  a  circumference."     (See  Prop.  XII.) 

551.  The  length  of  an  arc  of  a  circumference  is  such  a  part 
of  the  length  of  the  circumference  as  the  central  angle  which 
intercepts  the  arc  is  of  360°.     (See  §  360.) 

552.  The  approximate  length  of  a  circumference  is  found  in 
elementary  geometry  by  computing  the  perimeters  of  a  series 
of  regular  inscribed  and  circumscribed  polygons  which  are  ob- 
tained by  repeatedly  doubling  the  number  of  their  sides.     The 
perimeters   of    these   inscribed   and   circumscribed    polygons, 
since  they  approach  a  common  limit,  may  be  made  to  agree  to 
as  many  decimal  places  as  we  please,  according  to  the  number 
of  times  we  double  the  number  of  sides  of  the  polygons. 

PROPOSITION  XII.     THEOREM 

553.  The  ratio  of  the  circumference  of  a  circle  to  its 
diameter  is  the  same  for  all  circles. 


Given   any  two  circles  with  circumferences  (7  and  (/,  and  with 
radii  R  and  R',  respectively. 

To  prove   —  =  — . . 
2R      2R' 


BOOK   V 


265 


ARGUMENT 

1.    Inscribe   in   the   given   circles   regular 
polygons  of  the  same  number  of  sides, 
and  call  their  perimeters  .Pand  P'. 
P       R      2R 


2. 


p' 


4.  As  the  number  of  sides  of  the  two  regu- 

lar polygons  is  repeatedly  doubled, 
P  approaches  C  as  a  limit,  and  p' 
approaches  C1  as  a  limit. 

p  '    c1 

5.  .-.  —  approaches  —  as  a  limit. 

9;?  9  J? 

*-'  jt  *_  /t 

P'  CY' 

6.  Also  — |  approaches  t -  as  a  limit. 

p  p' 

7.  But  --is  always  equal  to  — -. 

2  R  2  R 

'  '2R~  2R*'  Q.E.D. 


REASONS 
1.  §  517,  a. 


2.  §  538. 

3.  §  396. 

4.  §  550. 


5.  §  408,  b. 

6.  §  408,  b. 

7.  Arg/3. 

8.  §355. 


554.  Def.     This  constant  ratio  of  the  circumference  of  a 
circle  to  its  diameter  is  usually  represented  by  the  Greek  let- 
ter  TT.     It   will  be   shown  (§  568)  that  its  value  is  approxi- 
mately 3-J-j  or,  more  accurately,  3.1416. 

555.  Cor.  I.    Tlxe  circumference  of  a  circleis  equal  to  2  *  R. 

556.  Cor.  II.    Any    two    circumferences    are    to   each 
other  as  their  radii.     


Ex.  1002.  If  the  radius  of  a  wheel  is  4  feet,  how  far  does  it  roll  in 
two  revolutions  ? 

Ex.  1003.  How  many  revolutions  are  made  by  a  wheel  whose  radius 
is  3  feet  in  rolling  44  yards  ? 

Ex.  1004.  (a)  Find  the  width  of  the  ring  between  two  concentric 
circumferences  whose  lengths  are  2  feet  and  3  feet,  respectively. 

(6)  Assuming  the  earth's  equator  to  be  25,000  miles,  find  the  width 
of  the  ring  between  it  and  a  concentric  circumference  1  foot  longer. 

(c)  Write  your  inference  in  the  form  of  a  general  statement. 


266 


PLANE   GEOMETRY 
PROPOSITION  XIII.     THEOREM 


557.   Tlie  area  of  a  regular  polygon  is  equal  to  one 
half  the  product  of  its  perimeter  and  its  apothem. 

C 


Given   regular  polygon  ABCD 
apothem. 

To  prove   area  of  ABCD  -  •  -  =  1  P  r. 

ARGUMENT 

1.  In  polygon  ABCD  •  •  • ,  inscribe  a  circle. 

2.  Then  r,  the  apothem  of  regular  polygon 

ABCD  •  •  • ,  is  the  radius  of  circle  0. 

3.  .-.  area  of  ABCD  •  •  •  =  1  P  r.        Q.E.D. 


,  P  its  perimeter,  and  r  its 


REASONS 
§  530. 

G  £33. 


3.    §  492. 


Ex.  1005.     Find  the  area  of  a  regular  hexagon  whose  side  is  0  inches. 

Ex.  1006.  The  area  of  an  inscribed  regular  hex- 
agon is  a  mean  proportional  between  the  areas  of  the 
inscribed  and  circumscribed  equilateral  triangles. 

Ex.  1007.  The  figure  represents  a  flower  bed 
drawn  to  the  scale  of  1  inch  to  20  feet.  Find  the 
number  of  square  feet  in  the  flower  bed. 


558.  Def.  The  area  of  a  circle  is  the  common  limit  which 
the  successive  areas  of  inscribed  and  circumscribed  regular 
polygons  approach  as  the  number  of  sides  is  successively  in- 
creased and  each  side  approaches  zero  as  a  limit. 


BOOK  V 


267 


PROPOSITION  XIV.     THEOREM 

559.    The  area  of  a  circle  is  equal  to  one  half  the  prod- 
uct of  its  circumference  and  its  radius. 


ircle  0,  with  radius  R,  circumference  C,  and  area  K. 


Given  cir 

To  prove    K  =     OR. 

ARGUMENT  REASONS 

1.  Circumscribe  about  circle  0  a  regular      1.    §  517,  b. 

polygon.      Call  its  perimeter  P  and 
its  area  S. 

2.  Thens  =  £p*.  2.   §  557. 

3.  As  the  number  of  sides  of  the  regular     3.    §  550. 

circumscribed  polygon  is  repeatedly 
doubled,  P  approaches  C  as  a  limit. 

4.  .*.  \PR  approaches  J  CR  as  a  limit.-  4.  §  561. 

5.  Also  S  approaches  .K"  as  a  limit.  5.  §. 

6.  But  S  is  always  equal  to  1  PR.  6.  Arg.  2. 

7.  ,'.K—\CR.  Q.E.D.  7.  §355. 

\ 

560.  TJie  product  of  a  variable  and  a  constant  is  a  variable. 

561.  The  limit  of  the  product  of  a  variable  omd  a  constant,  not 
zero,  is  the  limit  of  the  variable  multiplied  by  the  constant. 

(Proofs  of  these  theorems  will  be  found  in  the  Appendix,   §§  585  and 
590.) 

562.  Cor.  I.    The  area  of  a  circle  is  equal  to  irR~. 
HINT.     K=  \  C  •  R  =  |  •  2  irH  .  E  =  -rrT?2. 


268  PLANE   GEOMETRY 

563.  Cor.  II.    The  areas  of  two  circles  are  to  each  other 
as  the  squares  of  their  radii,  or  as  the  squares  of  their 
diameters. 

564.  Cor.  m.    The  area  of  a  sector  whose  angle  is  a°  is 

§  551.) 


Ex.  1008.    Find  the  area  of  a  circle  whose  radius  is  3  inches. 

Ex.  1009.  Find  the  area  of  a  sector  the  angle  of  which  is  (a)  45°, 
(6)  120°,  (c)  17°,  and  the  radius,  5  inches.  Find  the  area  of  the  segment 
corresponding  to  (&).  HINT.  Segment  =  sector  —  triangle. 

Ex.  1010.  If  the  area  of  one  circle  is  four  times  that  of  another,  and 
the  ra<f ius  of  the  first  is  6  inches,  what  is  the  radius  of  the  second  ? 

Ex.  1011.  Find  the  area  of  the  ring  included  between  the  circumfer- 
ences^of  two  concentric  circles  whose  radii  are  6  inches  and  8  inches. 

Ex.  1012.  Find  the  radius  of  a  circle 
whose  area  is  equal  to  the  sum  of  the  areas 
of  two  circles  with  radii  3  and  4,  respectively. 

Ex.  1013.  In  the  figure  the  diameter  AB 
=  2  R,  AC  =  \AB,  AD  =  i  AB,  and  E  is  any 
point  on  AB.  (1)  Find  arc  AC  +  arc  CB  ;  arc 
AD  +  arc  DB ;  arc  AE  -f  arc  EB.  (2)  Com- 
pare each  result  with  semicircumference  AB. 


565.  Def.  In  different  circles  similar  arcs,  similar  sectors, 
and  similar  segments  are  arcs,  sectors,  and  segments  that  cor- 
respond to  equal. central  angles. 


*  Ex.  1014.     Similar  arcs  are  to  each  other  as  their  radii. 

\        Ex.   1015.     Similar  sectors  are  to  each  other  as  the  squares  of  their 
iPadii. 

Ex.  1016.  Similar  seg- 
ments are  to  each  other  as 
thefttouares  of  their  radii. 

HINT.     Prove 
sector  A  OB     &AOB 


sector  CPD      A  CPD 

apply  §§  396  and  399.  M        ~  JY 


BOOK  V 


269 


PROPOSITION  XV.     PROBLEM 

566.  Given  a  circle  of  unit  diameter  and  the  side  of  a 
regular  inscribed  polygon  of  n  sides,  to  find  the  side  of 
a  regular  inscribed  polygon  of  2n  sides. 


Given  circle  ABF  of  unit  diameter,  AB  the  side  of  a  regular 
inscribed  polygon  of  n  sides,  and  CB  the  side  of  a  regular  in- 
scribed polygon  of  2  n  sides ;  denote  AB  by  s  and  CB  by  x. 

To  find   x  in  terms  of  s. 


ARGUMENT 

REASONS 

1. 

DraAv  diameter  CF  ;  draw  BO  and  BF. 

1. 

§  54,  15. 

2. 

Z  CBF  is  a 

rt.  Z. 

2. 

§367. 

3. 

Also  CF  is 

the  _L  bisector  of  AB. 

3. 

§142. 

4. 

.-.  as2  =  ci 

7  '  CK. 

4. 

§  443,  II. 

5. 

Now  C'/1  = 

1,  £0  =  -,  CO  =  -. 

o7                      o 

5. 

By  cons. 

6. 

o 

.-.  as  =  x2 

=  1  •  CK  =  CK  =  CO  —  KO 

6. 

§  309. 

=  2 

-KO. 

7. 

'"'aj2  =  2~'1 

\IBO"  —  Ini 

7. 

§447. 

1 

//1V     /sV    1  —  VI  —  s2 

2 

*  V-7      V2/              2 

8. 

-.*-xF 

-  VI  -s2 

8. 

§  54,  13. 

\ 

Q.E.F. 

270 


PLANE   GEOMETRY 


PROPOSITION  XVI.     PROBLEM 

567.  Given  a  circle  of  unit  diameter  and  the  side  of  a 
regular  circumscribed  polygon  of  n  sides,  to  find  the  side 
of  a  regular  circumscribed  polygon  of  2  n  sides. 


Given  circle  0  of  unit  diameter,  AB  half  the  side  of  a  reg- 
ular circumscribed  polygon  of  n  sides,  and  CB  half  the  side 
of  a  regular  circumscribed  polygon  of  2  n  sides;  denote  AB 

(j5bJ  f- 
find  x  in  terms  of  s. 


ARGUMENT 

1.  Draw  CO  and  AO. 

2.  Z  BOC=  |Z  BOA. 

3.  .'.  ill  A  OAB,  AC  :  CB  =  AO \  BO. 

4.  But     AC  =  AB  —  CB. 

5.  And    A0  = 


IAB  +  BO 

6.    .'.  AB  —  CB  :CB  — 


s  r 

7.    Substituting  -  for  AB,  ^  for  C'£,  and  - 


fcr^'|*|:.|- 


8. 


—  a:  :  x  =  Vs2  +  1:1. 


REASONS 

1.  §  54,  15. 

2.  §517,6. 

3.  §432. 

4.  §  54,  11. 

5.  §446. 
0.  §309. 
7.  §309. 


8.    §403. 


BOOK   V 


271 


ARGUMENT 

.  s  —  x  =  x  V 

•*  x  —  — 

REASONS 
9.    §388. 

10.    Solving  for  a;. 

o-       1        1 

8 

1  + 

Vs2  +  1 

Q.E.F. 

9. 
10. 


Ex.  1017.  Given  a  circle  of  unit  diameter  and  an  inscribed  and  a 
circumscribed  square  ;  compute  the  side  of  the  regular  inscribed  and  the 
regular  circumscribed  octagon. 


PROPOSITION  XVII.     PROBLEM 

568.  To  compute  the  approximate  value  of  the  circum- 
ference of  a  circle  in  terms  of  its  diameter ;  i.e.  to  compute 
the  value  of  v. 

B 


Given   circle  ABCD,  with  unit  diameter. 
To  compute   approximately  the  circumference  of  circle  ABCD 
in  terms  of  its  diameter;  i.e.  to  compute  the  value  of  TT. 


ARGUMENT 

1.  The   ratio   of   tne   circumference  of   a 

circle  to  its  d.araeter  is  the  same  for 
all  circles. 

2.  Since  the  diameter  of  the  given  circle  is 

unity,  the  side  of  an  inscribed  square 
will 'be  i  V£ 


REASONS 

1.  §553. 

2.  §522. 


272 


PLANE   GEOMETRY 


ARGUMENT 


3.  By  using  the  formula  x  =  A/-!-  ~  v  1  ~  ^y 

the  sides  of  regular  inscribed  polygons 
of  8,  16,  32,  etc.,  sides  may  be  com- 
puted ;  and  by  multiplying  the  length 
of  one  side  by  the  number  of  sides, 
the  length  of  the  perimeter  of  each 
'polygon  may  be  obtained.  The  re- 
sults are  given  in  the  table  below. 

4.  Likewise  if  the  diameter  of  the  given 

circle  is  unity,  the  side  of  a  circum- 
scribed square  will  be  1. 

5.  By  using  the  formula  x  = s  , 

1  +  Vs2  +  1 

the  sides  of  regular  circumscribed 
polygons  of  8,  16,  32,  etc.,  sides  may 
be  computed ;  and  by  multiplying  the 
length  of  one  side  by  the  number  of 
sides,  the  length  of  the  perimeter  of 
each  polygon  may  be  obtained.  The 
results  are  given  in  the  following  table. 


REASONS 


3.   §566. 


4.   §522. 


5.    §567. 


NUMBER  OP  SIDES 

PERIMETER  OF  INSCRIBED 

POLYGON 

PERIMETER  OF  CIRCUMSCRIBED 

POLYGON 

4 

2.828427 

4.000000 

8 

3.061467 

3.313708 

16 

3.121445 

3.182597 

32 

3.136548 

3.151724 

«r    64 

3.140331 

3.144118 

128 

3.141277 

3.142223 

256 

3.141513 

3.141750 

512 

3.141572    _ 

3.141632 

1054^ 

3.141587 

3.141602 

2048. 

3.141591 

3.141595 

4096 

3.141592 

3.141593 

BOOK  V  273 

These  successive  perimeters  will  be  closer  and  closer  approxi- 
mations of  the  length  of  the  circumference.  By  continuing  to 
double  the  number  of  sides  of  the  inscribed  and  circumscribed 
polygons,  perimeters  may  be  obtained  which  agree  to  as  many 
orders  of  decimals  as  desired. 

The  last  numbers  in  the  table  show  that  the  length  of  a  cir- 
cumference of  unit  diameter  lies  between  3.141592  and  3.141593. 

.-.  IT,  the  ratio  of  any  circumference  to  its  diameter,  to  five  deci- 
mal places  is  3.14159.  The  value  commonly  used  is  3.1416. 

569.  Historical  Note.  The  earliest  known  attempt  to  find  the  area 
of  a  circle  was  made  by  Ahmes,  an  Egyptian  priest,  as  early  as  1700  B.C. 
His  method  gave  for  TT  the  equivalent  of  3.1604.  His  manuscript  is  pre- 
served in  the  British  Museum. 

Archimedes  (250  B.C.)  gave  the  value  -2i^,  his  method  being  similar  to 
that  given  in  the  text. 

Hero  of  Alexandria  gave  3  and  3}. 

Ptolemy  (about  150  B.C.)  gave  3.1417. 

Metius  of  Holland  (1600  A.D.)  gave  fff ,  which  is  correct  to  six  places. 

Lambert  (1750  A.D.)  proved  TT  an  irrational  number,  and  Lindemann 
(1882)  proved  it  transcendental,  i.e.  not  expressible  as  a  root  of  an  alge- 
braic equation. 

By  methods  of  the  calculus  the  value  of  TT  has  been  computed  to  sev- 
eral hundred  places.  Richter  carried  it  to  500  decimal  places,  and  Shanks, 
in  1873,  gave  707  places. 

It  is  impossible  to  "  square  the  circle,"  i.e.  to  obtain  by  accurate  con- 
struction, with  the  use  of  ruler  and  compasses  only,  a  square  equivalent  to 
the  area  of  a  circle. 


Ex.  1018.     Find  the  area  of  a  circle  whose  radius  is  5  inches.     (Let 
r  =  8.1416.) 

Ex.  1019.     The  side  of  an  inscribed  square  is  4  inches.     What  is  the 
area  of  the  circle  ? 

Ex.  1020.  What  is  the  area  of  a  circle  inscribed 
in  a  square  whose  side  is  6  inches  ? 

Ex.  1021.  The  figure  represents  a  circular  grass 
plot  drawn  to  the  scale  of  1  inch  to  24  feet.  Measure 
carefully  the  radius  of  the  circle  and  find  the  number 
of  square  feet  in  the  grass  plot, 


274 


PLANE   GEOMETRY 


Ex.  1022.  The  figure  represents  a  belt  from  drive  wheel  O  to  wheel 
Q.  The  diameter  of  wheel  O  is 
2  feet  and  of  wheel  Q  16  inches. 
If  the  drive  wheel  makes  75  revo- 
lutions per  minute,  how  many  rev- 
olutions per  minute  will  the  smaller 
wheel  make  ? 


What  is  the  area  of  a 


Ex.  1023.     The  radius  of  a  circle  is  6  inches, 
segment  whose  arc  is  60°  ? 

Ex.  1024.  The  radius  of  a  circle  is  8  inches.  What  is  the  area  of 
the  segment  subtended  by  the  side  of  an  inscribed  equilateral  triangle  ? 

Ex.  1025.  The  diagonals  of  a  rhombus  are  16  and  30  ;  find  the  area 
of  the  circle  inscribed  in  the  rhombus. 

MISCELLANEOUS  EXERCISES 

Ex.  1026.  An  equiangular  polygon  inscribed  in  a  circle  is  regular  if 
the  number  of  sides  is  odd. 

Ex.  1027.  An  equilateral  polygon  circumscribed  about  a  circle  is 
regular  if  the  number  of  sides  is  odd. 

Ex.  1028.  Find  the  apothem  and  area  in  terms  of  the  radius  in  an 
equilateral  triangle  ;  in  a  square ;  in  a  regular  hexagon. 

Ex.  1029.  The  lines  joining  the  mid-points  of  the  apothems  of  a  regu- 
lar pentagon  form  a  regular  pentagon.  Find  the  ratio  of  its  area  to  the 
area  of  the  original  pentagon. 

Ex.  1030.  Within  a  circular  grass  plot  of  radius  6  feet,  a  flower  bed 
in  the  form  of  an  equilateral  triangle  is  inscribed.  How  many  square  feet 
of  turf  remain  ? 

Ex.  1031.  The  area  of  a  regular  hexagon  inscribed  in  a  circle  is  24  Vs. 
What  is  the  area  of  the  circle  ? 

Ex.  1032.  From  a  circle  of  radius  6  is  cut  a  sector  whose  central 
angle  is  105°.  Find  the  area  and  perimeter  of  the  sector.  (v  =  2r2.) 

Ex.  1033.  Prove  that  the  following  method  of 
inscribing  a  regular  pentagon  and  a  regular  decagon 
in  a  circle  is  correct.  Draw  diameter  CD  perpen- 
dicular to  diameter  AB ;  bisect  OA  and  join  its  mid- 
point to  D  ;  take  EF=  ED  and  draw  FD.  FD  will 
be  the  side  of  the  required  pentagon,  and  OF  the  side 
of  the  required  decagon. 


BOOK  V  275 

Ex.  1034.  Divide  a  given  circle  into  two  segments  such  that  any 
angle  inscribed  in  one  segment  is  twice  an  angle  inscribed  in  the  other ; 
so  that  an  angle  inscribed  in  one  segment  is  three  times  an  angle  inscribed 
in  the  other  ;  seven  times. 

Ex.  1035.  Show  how  to  cut  off  the  corners  of  an  equilateral  triangle 
so  as  to  leave  a  regular  hexagon  ;  of  a  square  to  leave  a  regular  octagon. 

Ex.  1036.  The  diagonals  of  a  regular  pentagon  form  a  regular  penta- 
gon. 

Ex.  1037.  The  diagonals  joining  alternate  vertices  of  a  regular  hexa- 
gon form  a  regular  hexagon  one  third  as  large  as  the  original  one. 

Ex.  1038.  The  area  of  a  regular  inscribed  octagon  is  equal  to  the 
product  of  the  side  of  an  inscribed  square  and  the  diameter. 

Ex.  1039.     If  a  is  the  side  of  a  regular  pentagon  inscribed  in  a  circle 

whose  radius  is  J?,  then     a  =  -^  v  10 —  2V5- 

'2  -  _ 

Ex.  1040.  The  area  of  a  regular  inscribed  dodecagon  is  equal  to  three 
times  the  square  of  the  radius. 

Ex.  1041.     Construct  an  angle  of  9°. 

Ex.  1042.     Construct  a  regular  pentagon,  given  one  of  the  diagonals. 

Ex.  1043.  Through  a  given  point  construct  a  line  which  shall  divide 
a  given  Circumference  into  two  parts  in  the  ratio  of  3  to  7  ;  in  the  ratio  of 
3  to  5.  Can  the  given  point  lie  within  the  circle? 

Ex.  1044.     Transform  a  given  regular  octagon  into  a  square. 

Ex.  1045.  Construct  a  circumference  equal  to  the  sum  of  two  given 
circumferences. 

Ex.  1046.  Divide  a  given  circle  by  concentric  circumferences  into 
four  equivalent  parts. 

Ex.  1047.  In  a  given  sector  whose  angle  is  a  right  angle  inscribe  a 
square. 

Ex.  1048.     In  a  given  sector  inscribe  a  circle. 

Ex.  1049.  If  two  chords  of  a  circle  are  perpendicular  to  each  other, 
the  sum  of  the  four  circles  having  the  four  segments  as  diameters  is 
equivalent  to  the  given  circle. 

Ex.  1050.  The  area  of  a  ring  between  two  concentric  circumferences 
whose  radii  are  E  and  R'  respectively  is  Tr(JP  —  E''2). 

Ex.  1051.  The  area  of  the  surf  ace 'between  two  concentric  circles  is 
equal  to  twice  the  area  of  the  smaller  circle.  Find  the  ratio  between 
their  radii. 


276  PLANE   GEOMETRY 

MISCELLANEOUS  EXERCISES   ON  PLANE  GEOMETRY 

Ex.  1052.  If  equilateral  triangles  are  constructed  on  the  sides  of 
any  given  triangle,  the  lines  joining  the  vertices  of  the  given  triangle  to 
the  outer  vertices  of  the  opposite  equilateral  triangles  are  equal. 

Ex.  1053.  If,  on  the  arms  of  a  right  triangle  as  diameters,  semicircles 
are  drawn  so  as  to  lie  outside  of  the  triangle,  and  if,  on  the  hypotenuse  as 
a  diameter,  a  semicircle  is  drawn  passing  through  the  vertex  of  the  right 
angle,  the  sum  of  the  areas  of  the  two  crescents  included  between  the 
semicircles  is  equal  to  the  area  of  the  given  triangle. 

Ex.  1054.  The  area  of  the  regular  inscribed  triangle  is  half  that  of 
the  regular  inscribed  hexagon. 

Ex.  1055.  From  a  given  point  draw  a  secant  to  a  circle  such  that  its 
internal  and  external  segments  shall  be  equal. 

Ex.  1056.  Show  that  the  diagonals  of  any  quadrilateral  inscribed  in 
a  circle  diyide  the  quadrilateral  into  four  triangles  which  are  similar,  two 
and  two. 

Ex.  1057.  Through  a  point  P,  outside  of  a  circle,  construct  a  secant 
PAB  so  that  AT?  =  PA  x  PB. 

Ex.  1058.  The  radius  of  a  circle  is  6  feet.  What  are  the  radii  of  the 
circles  concentric  with  it  whose  circumferences  divide  its  area  into  three 
equivalent  parts  ? 

» f  C 

Ex.  1059.     Given  parallelogram  ABCD, 
F\hQ  mid-point  of  BC ;  prove  OT=\TC. 

Ex.  1060.     Given  FT  a  tangent  to  a  circle     J* 
at  point  T,  and  two  other  tangents  parallel 

to  each  other  cutting  PT  at  A  and  B  respectively  ;  prove  that  the  radius 
of  the  circle  is  a  mean  proportional  between  .AT7 and  TB. 

Ex.  1061.  Show  that  a  mean  proportional  between  two  unequal  lines 
is  less  than  half  their  sum. 

Ex.  1062.  Given  two  similar  triangles,  construct  a  triangle  equiva- 
lent to  their  sum. 

Ex.  1063.  The  square  of  the  side  of  an  inscribed  equilateral  triangle 
is  equal  to  the  sum  of  the  squares  of  the  sides  of  the  inscribed  square  and 
of  the  inscribed  regular  hexagon. 

Ex.  1064.  Prove  that  the  area  of  a  circular  ring  is  equal  to  the  area 
of  a  circle  whose  diameter  equals  a  chord  of  the  outer  circumference  which 
is  tangent  to  the  inner. 


BOOK  V  277 

Ex.  1065.  It  two  chords  drawn  from  a  common  point  P  on  the  cir- 
cumference of  a  circle  are  cut  by  a  line  parallel  to  the  tangent  through  P, 
the  chords  and  the  segments  of  the  chords  between  the  two  parallel  lines 
are  inversely  proportional. 

Ex.  1066.  Construct  a  segment  of  a  circle  similar  to  two  given  simi- 
lar segments  and  equivalent  to  their  sum. 

Ex.  1067.  The  distance  between  two  parallels  is  a,  and  the  distance 
between  two  points  A  and  B  in  one  parallel  is  2  b.  Find  the  radius  of 
the  circle  which  passes  through  A  and  B,  and  is  tangent  to  the  other 
parallel. 

Ex.  1068.  Tangents  are  drawn  thrqugh  a  point  6  inches  from  the 
circumference  of  a  circle  whose  radius  is  9  inches.  Find  the  length  of 
the  tangents  and  also  the  length  of  the  chord  joining  the  points  of  contact. 

Ex.  1069.  If  the  perimeter  of  each  of  the  figures,  equilateral  triangle, 
square,  and  circle,  is  396  feet,  what  is  the  area  of  each  figure  ? 

Ex.  1070.  The  lengths  of  two  sides  of  a  triangle  are  13  and  15  inches, 
and  the  altitude  on  the  third  side  is  12  inches.  Find  the  third  side,  and 
also  the  area  of  the  triangle.  (Give  one  solution  only.) 

Ex.  1071.  If  the  diameter  of  a  circle  is  3  inches,  what  is  the  length 
of  an  arc  of  80°  ? 

Ex.  1072.  AD  and  BC  are  the  parallel  sides  of  a  trapezoid  ABCD, 
whose  diagonals  intersect  at  E.  If  F  is  the  mid-point  of  BC,  prove  that 
FE  prolonged  bisects  AD. 

Ex.  1073.  Given  a  square  ABCD.  Let  E  be  the  mid-point  of 
CD,  and  draw  BE.  A  line  is  drawn  parallel  to  BE  and  cutting  the 
square.  Let  P  be  the  mid-point  of  the  segment  of  this  line  within  the 
square.  Find  the  locus  of  P  when  the  line  moves,  always  remaining 
parallel  to  BE.  Describe  the  locus  exactly,  and  prove  the  correctness  of 
your  answer. 

Ex.  1074.  Let  ABCD  be  any  parallelogram,  and  from  any  point  P 
in  the  diagonal  AC  draw  the  straight  line  PM  cutting  AB  in  M,  BC  in  JV, 
CD  in  L,  and  AD  in  K.  Prove  that  PM  •  P7V=  PK  -  PL. 

Ex.  1075.  Find  the  area  of  a  segment  of  a  circle  whose  height  is 
4  inches  and  chord  8\/3  inches. 

Ex.  1076.  A  square,  whose  side  is  5  inches  long,  has  its  corners 
cut  off  in  such  a  way  as  to  make  it  into  a  regular  octagon.  Find  the  area 
and  the  perimeter  of  the  octagon. 

Ex.  1077.  Into  what  numbers  of  arcs  less  than  15  can  the  circumfer- 
ence of  a  circle  be  divided  with  ruler  and  compasses  only  ? 


278  PLANE   GEOMETRY 

Ex.  1078.  Through  a  point  A  on  the  circumference  of  a  circle  chords 
are  drawn.  On  each  one  of  these  chords  a  point  is  taken  one  third  of  the 
distance  from  A  to  the  other  end  of  the  chord.  Find  the  locus  of  these 
points,  and  prove  that  your  answer  is  correct. 

Ex.  1079.  In  what  class  of  triangles  do  the  altitudes  meet  within  the 
triangle  ?  on  the  boundary  ?  outside  the  triangle  ?  Prove. 

Ex.  1080.  Given  a  triangle  ABC  and  a  fixed  point  D  on  side  AC', 
draw  the  line  through  D  which  divides  the  triangle  into  two  parts  of 
equal  area. 

Ex.  1081.  The  sides  of  a  triangle  are  5,  12,  13.  Find  the  radius  of 
the  circle  whose  area  is  equal  to  that  of  the  triangle. 

Ex.  1082.  In  a  triangle  ABC  the  angle  C  is  a  right  angle,  and  the 
lengths  of  AC  and  BC  are  6  and  12  respectively  ;  the  hypotenuse  BA  is 
prolonged  through  A  to  a  point  D  so  that  the  length  of  AD  is  4  ;  CA  is 
prolonged  through  A  to  E  so  that  the  triangles  AED  and  ABC  have 
equal  areas.  What  is  the  length  of  AE? 

Ex.  1083.  Given  three  points  A,  B,  and  (7,  not  in  the  same  straight 
line ;  through  A  draw  a  straight  line  such  that  the  distances  of  B  and  C 
from  the  line  shall  be  equal. 

Ex,  1084.  Given  two  straight  lines  that  cut  each  other ;  draw  four 
circles  of  given  radius  that  shall  be  tangent  to  both  of  these  lines. 

Ex.  1085.  Construct  two  straight  lines  whose  lengths  are  in  the  ratio 
of  the  areas  of  two  given  polygons. 

Ex.  1086.  The  radius  of  a  regular  inscribed  polygon  is  a  mean  pro- 
portional between  its  apothem  and  the  radius  of  the  similar  circumscribed 
polygon. 

Ex.  1087.  Draw  a  circumference  which  shall  pass  through  two  given 
points  and  bisect  a  given  circumference. 

Ex.  1088.  A  parallelogram  is  constructed  having  its  sides  equal  and 
parallel  to  the  diagonals  of  a  given  parallelogram.  Show  that  its  diagonals 
are  parallel  to  the  sides  of  the  given  parallelogram. 

HINT.    Look  for  similar  triangles. 

Ex.  1089.  If  two  chords  are  divided  in  the  same  ratio  at  their  point 
of  intersection,  the  chords  are  equal. 

Ex.  1090.  The  sides  AB  and  AC  of  a  triangle  ABC  are  bisected  in 
D  and  E  respectively.  Prove  that  the  area  of  the  triangle  DBG  is  twice 
that  of  the  triangle  DEB. 

Ex.  1091.  Two  circles  touch  externally.  How  many  common  tan- 
gents have  they  ?  Give  a  construction  for  the  common  tangents. 


BOOK  V  279 

Ex.  1092.  Prove  that  the  tangents  at  the  extremities  of  a  chord  of  a 
circle  are  equally  inclined  to  the  chord. 

Ex.  1093.  Two  unequal  circles  touch  externally  at  P ;  line  AB  touches 
the  circles  at  A  and  B  respectively.  Prove  angle  APB  a  right  angle. 

Ex.  1094.  Find  a  point  within  a  triangle  such  that  the  lines  joining 
this  point  to  the  vertices  shall  divide  the  triangle  into  three  equivalent 
parts. 

Ex.  1095.  A  triangle  ABC  is  inscribed  in  a  circle.  The  angle  B  is 
equal  to  60°  and  the  angle  G  is  equal  to  60°.  What  angle  does  a  tangent 
at  A  make  with  BC  prolonged  to  meet  it  ? 

Ex.  1096.  The  bases  of  a  trapezoid  are  8  and  12,  and  the  altitude  is 
6.  Find  the  altitudes  of  the  two  triangles  formed  by  prolonging  the  non- 
parallel  sides  until  they  intersect. 

Ex.  1097.  The, circumferences  of  two  circles  intersect  in  the  points  A 
and  B.  Through  A  a  diameter  of  each  circle  is  drawn,  viz.  AC  and  AD. 
Prove  that  the  straight  line  joining  C  and  D  passes  through  B. 

Ex.  1098.  How  many  lines  can  be  drawn  through  a  given  point  in  a 
plane  so  as  to  form  in  each  case  an  isosceles  triangle  with  two  given  lines 
in  the  plane  ? 

Ex.  1099.  The  lengths  of  two  chords  drawn  from  the  same  point  in 
the  circumference  of  a  circle  to  the  extremities  of  a  diameter  are  5  feet 
and  12  feet  respectively.  Find  the  area  of  the  circle. 

Ex.  1100.  Through  a  point  21  inches  from  the  center  of  a  circle 
whose  radius  is  15  inches  a  secant  is  drawn.  Find  the  product  of  the 
whole  secant  and  its  external  segment. 

Ex.  1101.  The  diagonals  of  a  rhombus  are  24  feet  and  40  feet  re- 
spectively. Compute  its  area. 

Ex.  1102.  On  the  sides  AB,  BC,  CA  of  an  equilateral  triangle  ABC 
measure  off  segments  AD,  BE,  CF,  respectively,  each  equal  to  one  third 
the  length  of  a  side  ;  draw  triangle  DEF ;  prove  that  the  sides  of  triangle 
DEF3.T&  perpendicular  respectively  to  the  sides  of  triangle  ABC. 

Ex.1103.     Construct  x  if  (a)  ->  =  ^ ;    (b)x  =  aVb. 

Ex.  1104.  Find  the  area  included  between  a  circumference  of 
radius  7  and  an  inscribed  square. 

Ex.  1105.  What  is  the  locus  of  the  center  of  a  circle  of  given 
radius  whose  circumference  cuts  at  right  angles  a  given  circumference  ? 

Ex.  1106.  Two  chords  of  a  certain  circle  bisect  each  other..  One  of 
them  is  10  inches  long ;  how  far  is  it  from  the  center  of  the  circle  ? 


280  PLANE   GEOMETRY 

Ex.  1107.  Show  how  to  find  on  a  given  straight  line  of  indefinite 
length  a  point  0  which  shall  be  equidistant  from  two  given  points  A  and 
B  in  the  plane.  If  A  and  E  lie  on  a  straight  line  which  cuts  the  given 
line  at  an  angle  of  45°  at  a  point  7  inches  distant  from  A  and  17  inches 
from  J5,  show  that  OA  will  be  13  inches. 

Ex.  1108.  A  variable  chord  passes,  when  prolonged,  through  a 
fixed  point  outside  of  a  given  circle.  What  is  the  locus  of  the  mid-point 
of  the  chord  ? 

Ex.  1109.  A  certain  parallelogram  inscribed  in  a  circle  has  two 
sides  20  feet  in  length  and  two  sides  15  feet  in  length.  What  are  the 
lengths  of  the  diagonals  ? 

Ex.  1110.  Upon  a  given  base  is  constructed  a  triangle  one  of  the 
base  angles  of  which  is  double  the  other.  The  bisector  of  the  larger  base 
angle  meets  the  opposite  side  at  the  point  P.  Find  the  locus  of  P. 

Ex.  1111.  What  is  the  locus  of  the  point  of  contact  of  tangents 
drawn  from  a  iixed  point  to  the  different  members  of  a  system  of  con- 
centric circles  ? 

Ex.  1112.  Find  the  locus  of  all  points,  the  perpendicular  distances 
of  which  from  two  intersecting  lines  are  to  each  other  as  3  to  2. 

Ex.  1113.  The  sides  of  a  triangle  are  a,  6,  c.  Find  the  lengths  of 
the  three  medians. 

Ex.  1114.  Given  two  triangles;  construct  a  square  equivalent  to 
their  sum. 

Ex.  1115.  In  a  circle  whose  radius  is  10  feet,  two  parallel  chords 
are  drawn,  each  equal  to  the  radius.  Find  the  area  of  the  portion  be- 
tween these  chords. 

Ex.  1116.  A  has  a  circular  garden  and  B  one  that  is  square.  The 
distance  around  each  is  the  same,  namely,  120  rods.  Which  has  the 
more  land,  A  or  B  ?  How  much  more  has  he  ? 

Ex.  1117.  Prove  that  the  sum  of  the  angles  of  a  pentagram  (a  five- 
pointed  star)  is  equal  to  two  right  angles. 

Ex.  1118.  AB  and  A' B'  are  any  two  chords  of  the  outer  of  two  con- 
centric circles  ;  these  chords  intersect  the  circumference  of  the  inner  circle 
in  points  P,  Q  and  P',  Q'  respectively  :  prove  that^lP.  PB=A'P'  •  P'B'. 

Ex.  1119.  A  running  track  consists  of  two  parallel  straight  por- 
tions joined  together  at  the  ends  by  semicircles.  The  extreme  length  of 
the  plot  inclosed  by  the  track  is  176  yards.  If  the  inside  line  of  the 
track  is  a  quarter  of  a  mile  in  length,  find  the  cost  of  seeding  this  plot  at 
\  cent  a  square  yard.  (TT  —  -2r4.) 


BOOK  V  281 

Ex.  1120.  If  two  similar  triangles,  ABC  and  DEF,  have  their 
homologous  sides  parallel,  the  lines  AD,  BE,  and  CF,  which  join  their 
homologous  vertices,  meet  in  a  point. 

Ex.  1121.  In  an  acute  triangle  side  AB  =  10,  AC=T,  and  the  pro- 
jection of  AC  on  AB  is  3.4.  Construct  the  triangle  and  compute  the  third 
side  BC. 

Ex.  1122.  Divide  the  circumference  of  a  circle  into  three  parts  that 
shall  be  in  the  ratio  of  1  to  2  to  3. 

Ex.  1123.  The  circles  having  two  sides  of  a  triangle  as  diameters 
intersect  on  the  third  side. 

Ex.  1124.  Construct  a  circle  equivalent  to  the  sum  of  two  given 
circles. 

Ex.  1125!  Assuming  that  the  areas  of  two  triangles  which  have  an 
angle  of  the  one  equal  to  an  angle  of  the  other  are  to  each  other  as  the 
products  of  the  sides  including  the  equal  angles,  prove  that  the  bisector 
of  an  angle  of  a  triangle  divides  the  opposite  side  into  segments  propor- 
tional to  the  adjacent  sides. 

Ex.  1126.  In  a  circle  of  radius  5  a  regular  hexagon  is  inscribed. 
Determine  (a)  the  area  of  one  of  the  segments  of  the  circle  which  are 
exterior  to  the  hexagon; '(6)  the  area  of  a  triangle  whose  vertices  are 
three  successive  vertices  of  the  hexagon  ;  (c)  the  area  of  the  ring  bounded 
by  the  circumference  of  the  given  circle  and  that  of  the  circle  inscribed 
in  the  hexagon. 

Ex.  1127.  Find  the  locus  of  the  extremities  of  tangents  to  a  given 
circle,  which  have  a  given  length. 

Ex.  1128.  A  ladder  rests  with  one  end  against  a  vertical  wall  and 
the  other  end  upon  a  horizontal  floor.  If  the  ladder  falls  by  sliding 
along  the  floor,  what  is  the  locus  of  its  middle  point  ? 

Ex.  1129.  An  angle  moves  so  that  its  magnitude  remains  constant 
and  its  sides  pass  through  two  fixed  points.  Find  the  locus  of  the  vertex. 

Ex.  1130.  The  lines  joining  the  feet  of  the  altitudes  of  a  triangle 
form  a  triangle  whose  angles  are  bisected  by  the  altitudes. 

Ex.  1131.     Construct  a  triangle,  given  the  feet  of  the  three  altitudes. 

Ex.  1132.  If  the  radius  of  a  sector  is  2,  what  is  the  area  of  a  sector 
whose  central  angle  is  152°  ? 

Ex.  1133.  The  rectangle  of  two  lines  is  a  mean  proportional  between 
the  squares  on  the  lines. 

Ex.  1134.  Show  how  to  inscribe  in  a  given  circle  a  regular  polygon 
similar  to  a  given  regular  polygon. 


282 


PLANE   GEOMETRY 


FORMULAS   OF   PLANE   GEOMETRY 

570.   In  addition  to  the  notation  given  in  §  270,  the  follow- 
ing will  be  used : 


a  =  side  of  polygon  in  general. 
b  =  base  of  a  plane  figure. 
&,  6'  —  bases  of  a  trapezoid. 
C  =  circumference  of  a  circle. 
D'=  diameter  of  a  circle. 
E  =  sum  of  exterior  angles  of  a 

polygon. 

h  =  altitude  of  a  plane  figure. 
/  =  suin  of  interior  angles  of  a 

polygon. 

K  =  area  of  a  figure  in  general. 
I  =  line  in  general. 
P=  peri  meter    of    polygon    in 
general. 


p  =  projection  of  6  upon  a. 

_R  =  radius  of  circle,  or  radius  of 

regular  polygon. 

r  =  apothem   of  regular  polygon, 

or  radius  of  inscribed  circle. 

s  =  the  longer  of  two  segments  of 

a  line ;  or 
8  =  \(a  +  b  +  c). 
X  =  angle  in  general. 
xa  =  side  of  a  triangle  opposite  an 

acute  angle. 

x<>  =  side  of  a  triangle  opposite  an 
obtuse  angle. 


FIGURE 

Any  triangle. 
Polygon. 

Central  angle. 

Inscribed  angle. 

Angle  formed  by  two 
chords. 

Angle  formed  by  tangent 
and  chord. 

Angle  formed  by  two 
secants. 

Angle  formed  by  secant 
and  tangent. 

Angle  formed  by  two  tan- 
gents. 

Similar  polygons. 

Right  triangle. 
Any  triangle. 
Obtuse  triangle. 


FORMULA 

REFERENCE 

A+  B+  0=180°. 

§204. 

1=  (n  —  2)180°/i<V 

§§  216,  219. 

E  =  4  rt.  A. 

§218. 

X&  intercepted  arc. 

§  358. 

X  <*  i  intercepted  arc. 

§  3(55. 

JT°c  i  sum  of  arcs. 

§  377. 

X<£\  intercepted  arc. 
X  cc  i  difference  of  arcs. 
JT5?  \  difference  of  arcs. 
X^  I  difference  of  arcs. 

P  =  a_ 
P'     a' 


-2  ap. 
=  a2  +  62  +  2  ap. 


§378. 

§379. 

§379. 

.    §379. 

§  441. 

§446. 
§452. 
§455. 


FORMULAS 


283 


FIGURE 

FORMULA                            REFERENCE 

Any  triangle. 

P+*. 

2(^  +  2  w(l2.                          §457. 

Line    divided   in    extreme 

l:s  = 

s  :  I  -  s.               §  465  and  Ex.  763. 

and  mean  ratio. 

'.   Rectangle. 

K  = 

b  ;  -  h.                                        §  475. 

Square. 

K- 

a2.                                             §  478. 

Parallelogram. 

K  = 

'6  •  h.                                          §  481. 

V  Triangle.  A 

K  = 

1  b  -  h.                                       §  485. 

L        & 

ha  = 

^ 

^Vs(s  -  a)  (s  -  b)  (s  -  c).     §  490. 

.a 

•     , 

Jg'  =  Vs(s-«)(s-6)(s-c).       §490. 

* 

K  = 

i(a  +  b  +  c)r.                          §  491. 

Polygon. 

K  = 

iP-r.                                        §492. 

Circle  inscribed  in  tjritinorle 

Vs(s  —  a)(s—  b~)(s—  c)        Fx   g<J7 

Circle  circumscribed  about 

s 
abc                       Ex  838 

triangle.  K^ 

4v  s(s  —  ft)  (s  —  ?>)  (s  —  c) 

Bisector  of    angle  of  tri- 
angle. 

ta- 

o 

•\/7i/io/'o          /v\                          TTv     Q41 

6  +  C 

Trapezoid.  ;, 

K  = 

|(fi  +  ft')'*-                               §  495. 

Regular  polygons  of  same 

P  _ 

E      r 

number  of  sides. 

^,=V'                         §538- 

K_ 

^2  =  ^2                                                     §539' 

Circles. 

_d_  = 

r"      1 
2^'                                           §558' 

c  = 

2  TT.B.                                         §  555. 

c__ 

•p                                                               * 
J-it                                                          (\     +                                C     K  CP 

^/'                   /o                 *     * 

Regular  polygon. 

K  = 

|P-r.     N<L^N/v  ^^x            §557. 

Circle. 

K  = 

IC-BT      x  ,^     rJ"            §559> 
irT?2.      \^     ^  X^"                     §  562' 

Circles. 

K  _ 

^  =  ^.                               '§563. 

K1 

R»*      D'2 

oector. 

if 

central  Z  ^^                           ~  ^ 

j\.  — 

360D 

Segment. 

K  = 

sector  ^  triangle.             Ex.  1009. 

APPENDIX   TO   PLANE   GEOMETRY 

MAXIMA  AND  MINIMA 
I 

571.  Def.     Of  all  geometric  magnitudes  that  satisfy  given 
conditions,  the  greatest  is  called  the  maximum.,  and  the  least 
is  called  the  minimum.* 

572.  Def.     Isoperimetric  figures  are  figures  which  have  the 
same  perimeter. 

PROPOSITION  I.     THEOREM 

573.  Of  all  triangles  having  two  given  sides,  that  in 
which  these  sides  include  a  right  angle  is  the  maximum. 

B 


C 

Given  A  ABC  and  AEC,  with  AB  and  AC  equal  to  AE  and  AC 
respectively.     Let  Z  CAB  be  a  rt.  Z.  and  /.  CAE  an  oblique  /.. 
To  prove  A  ABC  >  A  AEC. 
Draw  the  altitude  EF. 
A  ABC  and  AEC  have  the  same  base,  AC. 
Altitude  AB  >  altitude  EF. 

.'.  A  ABC  >  A  AEC.  Q.E.D. 

574.  Cor.  I.  Conversely,  if  two  sides  are  given,  and  if 
the  triangle  is  a  maximum,  then  the  given  sides  include 
a  right  angle. 

HINT.     Prove  by  reductio  ad  absurdum. 
*  III  totfl:  mathematics  a  §9W§what  bro 


APPENDIX 


285 


575.  Cor.  II.     Of  all  parallelograms  Tiaving  given  sides, 
the  one  that  is  rectangular  is  a  maximum,  and  con- 
versely. 

Ex.  1135.  Construct  the  maximum  parallelogram  having  two  lines  of 
given  lengths  as  diagonals. 

Ex.  1136.  What  is  the  minimum  line  from  a  given  point  to  a  given 
line? 

Ex.  1137.  Of  all  triangles  having  the  same  base  and  altitude,  that 
which  is  isosceles  has  the  minimum  perimeter. 

PROPOSITION  II.     THEOREM 

576.  Of  all  equivalent  triangles  having  the  same  base, 
that  which  is  isosceles  has  the  least  perimeter. 

G 

A 


A  C 

Given  equivalent  A  ABC  and  AKC  with  the  same  base  AC, 
and  let  AB  =  BG  and  AE^EC. 

To  prove    AB  +  BC  +  CA  <  AE  +  EC  +  CA. 

Draw  CFA.AC  and  let  CF  meet  the  prolongation  of  AB  at  G. 

Draw  EG  and  BE  and  prolong  BE  to  meet  GC  at  F. 

BF  II  AC. 

Z  CBF  =  ^FBG. 

BF  bisects  CG  and  is  _L  CG. 

.-,  BG  =  BG  and  EC  =  EG. 

AB  +  BG  <  AE  +  EG. 
.'.  AB  +  BC  <  AE-\-EC. 
.-.  AB  +  BC  +  CA  <  AE  +  #G' 4-  CA.  Q.E.D. 


286  PLANE   GEOMETRY 

Ex.  1138.  Of  all  equivalent  triangles  having  the  same  base,  that 
which  has  the  least  perimeter  is  isosceles.  (Prove  by  reductio  ad  absur- 
dum.} 

Ex.  1139.  Of  all  equivalent  triangles,  the  one  that  has  the  minimum 
perimeter  is  equilateral. 

Ex.  1140.     State  and  prove  the  converse  of  Ex.  1139. 

PROPOSITION  III.     THEOREM 

577.  Of  all  isoperimetric  triangles  on  the  same  base, 
the  isosceles  triangle  is  the  maximum. 


A        .  H  C 

Given   isosceles  A  ABC  and  any  other  A  as  AEC  having  the 
same  base  and  the  same  perimeter  as  A  ABC. 
To  prove    A  ABC  >  A  AEC. 
Draw  BH _L  AC,  EF  from  E  II  AC,  and  draw  AF  and  FC. 

A  AFC  is  isosceles. 

.-.  perimeter  of  A  AFC  <  perimeter  of  A  AEC. 

.\  perimeter  of  A  AFC  <  perimeter  of  A  ABC. 

.'.  AF  <  AB. 

.'.  FH  <  BH. 

.'.  A  AFC  <  A  ABC. 

.-.  A  AEC  <  A  ABC',  i.e.  A  ABC  >  A  AEC.  Q.E.D. 


Ex.  1141.  Of  all  triangles  having  a  given  perimeter  and  a  given  base, 
the  one  that  has  the  maximum  area  is  isosceles. 

Ex.  1142.  What  is  the  maximum  chord  of  a  circle  ?  What  is  the 
maximum  and  what  the  minimum  line  that  can  be  drawn  from  a  given 
exterior  point  to  a  given  circumference  ? 


APPENDIX  287 

Ex.  1143.  Of  all  triangles  having  a  given  perimeter,  the  one  that 
has  the  maximum  area  is  equilateral. 

PROPOSITION  IV.     THEOREM 

578.  Of  all  polygons  having  all  their  sides  but  one 
equal,  respectively,  to  given  lines  taken  in  order,  the 
maximum  can  be  inscribed  in  a  semicircle  having  tlw 
undetermined  side  as  diameter. 

C 

E 


A  H 

Given  polygon  ABCEFH,  the  maximum  of  all  polygons  sub- 
ject to  the  condition  that  AB,  BC,  CE,  EF,  FH,  are  equal  respec- 
tively to  given  lines  taken  in  order. 

To  prove  that  the  semicircumference  described  with  AH  as 
diameter  passes  through  B,  C,  E,  and  F. 

Suppose  that  the  semicircumference  with  AH  as  diameter 
does  not  pass  through  some  vertex,  as  E.  Draw  AE  and  EH. 

Then  Z  AEH  is  not  a  rt.  Z. 

Then  if  the  figures  ^.BC^and  EFHaxe  revolved  about  E  until 
AEH  becomes  a  rt.  Z,  A  AEH  will  be  increased  in  area. 

.•.  polygon  ABCEFHc&n  be  increased  in  area  without  chang- 
ing any  of  the  given  sides. 

But  this  contradicts  the  hypothesis  that  polygon  ABCEFH  is 
a  maximum. 

/.  the  supposition  that  vertex  E  is  not  on  the  semicircumfer- 
ence is  false. 

.-.  the  semicircumference  passes  through  E. 

In  the  same  way  it  may  be  proved  that  every  vertex  of  the 
polygon  lies  on  the  semicircumference.  Q.E.D. 


Ex.  1144.     Given  the  base  and  the  vertex  angle  of  a  triangle,  con- 
struct the  triangle  so  that  its  area  shall  be  a  maximum. 


288 


PLANE    GEOMETRY 


Ex.  1145.  Find  the  point  in  a  given  straight  line  such  that  the  tan- 
gents drawn  from  it  to  a  given  circle  contain  a  maximum  angle. 

PROPOSITION  V.     THEOREM 

579.  Of  all  polygons  that  have  their  sides  equal,  re- 
spectively, to  given  lines  taken  in  order,  the  polygon 
that  can  be  circumscribed  by  a  circle  is  a  maximum. 


A'/ 


Given  polygon  ABCD  which  is  circumscribed  by  a  O,  and 
polygon  A'B'C'D'  which  cannot  be  circumscribed  by  a  O,  with 
AB  =  A'B',  BC  =  B'C',  CD=C'D',  and  DA  =  D' A'. 

To  prove    ABCD  >  A'B'c'D1. 

From  any  vertex  as  A  draw  diameter  AE\  draw  EC  and  ED. 
On  C'D',  which  equals  CD,  construct  AD'C'E'  equal  to  ADCE-, 
draw  A'E'. 

The  circle  whose  diameter  is  A'E'  does  not  pass  through  all 
the  points  B1,  C1,  D'.  (Hyp.) 

.-.  either  ABCE  or  EDA  or  both  must  be  greater,  and  neither 
can  be  less,  than  the  corresponding  part  of  polygon  A'B'C'E'D' 
(§  578). 

.-.  ABCED  >  A'B'C'E'D'.    But  A  DCE  =  A  D'C'E'. 
.-.  ABCD  >  A'B'C'D'.  Q.E.D. 


Ex.  1146.  In  a  given  semicircle  inscribe  a  trapezoid  whose  area  is  a 
maximum. 

Ex.  1147.  Of  all  equilateral  polygons  having  a  given  side  and  a  given 
number  of  sides,  the  one  that  is  regular  is  a  maximum. 


APPENDIX  289 


PROPOSITION  VI.     THEOREM 

580.    Of  all  isoperimetric  polygons  of  the  same  number 
of  sides,  the  maximum  is  equilateral. 


Given   polygon   ABODE  the  maximum  of   all    isoperimetric 
polygons  of  the  same  number  of  sides. 

To  prove    AB  =  B  C  =  CD  =  DE  =  EA. 

Suppose,  if  possible,  EG  >  CD. 

On  BD  as  base  construct  an  isosceles  A  BFD  isoperimetric 
with  A  BCD. 

ABFD>ABCD. 

.-.  polygon  ABFDE  >  polygon  ABODE. 

But  this  contradicts  the  hypothesis  that  ABODE  is  the  maxi- 
mum of  all1  isoperimetric  polygons  having  the  same  number  of 

sides. 

.-.  BC  =  CD. 

In  like  manner  any  two  adjacent  sides  may  be  proved  equal. 
.-.  AB  =  BO  =  CD  =  DE  ==  EA.  Q.E.D. 

581.   Cor.     Of  all  isoperimetric  polygons  of  the  same 
number  of  sides,  the  maximum  is  regular. 


Ex.  1148.     In  a  given  segment  inscribe  a  triangle  whose  perimeter  is 
a  maximum, 


290 


PLANE   GEOMETRY 


PROPOSITION  VII.     THEOREM 

582.    Of  two  isoperiinetric  regular  polygons,  that  which 
has  the  greater  number  of  sides  has  tlie  greater  area. 

F 
C 


H 


E 


Given  the  isoperimetric  polygons  P  and  Q,  and  let  P  have  one 
more  side  than  Q. 

To  prove  P  >  Q. 

In  one  side  of  Q  take  any  point  as  H. 

EFHG  may  be  considered  as  an  irregular  polygon  having  the 
same  number  of  sides  as  P. 

.-.  P  >  EFHG ;   i.e.  P  >  Q.  Q.E.D. 

PROPOSITION  VIII.     THEOREM 

583.  Of  two  equivalent  regular  polygons,  that  which 
has  the  greater  number  of  sides  has  the  smaller  pe- 
rimeter. 


H 


Given  square  S  =c=  regular  hexagon  H. 

To  prove  perimeter  of  S  >  perimeter  of  H. 

Construct  square  R  isoperimetric  with  H. 


APPENDIX  291 

Area  of  H  >  area  of  R ;  i.e.  area  of  s  >  area  of  R. 
.-.  perimeter  of  S  >  perimeter  of  R. 
.-.  perimeter  of  S  >  perimeter  of  H.  Q.E.D. 

584.  Cor.     Of  all  polygons  having  a  given  number  of 
sides  and  a  given   area,  that  which   has   a  minimum 
perimeter  is  regular. 

Ex.  1149.  Among  the  triangles  inscribed  in  a  given  circle,  the  one 
that  has  a  maximum  perimeter  is  equilateral. 

Ex.  1150.  Of  all  polygons  having  a  given  number  of  sides  and  in- 
scribed in  a  given  circle,  the  one  that  has  a  maximum  perimeter  is  regular. 

VARIABLES   AND   LIMITS.     THEOREMS 
PROPOSITION  I.     THEOREM 

585.  If  a  variable  can  be  made  less  than  any  assigned 
value,  the  product  of  the  variable  and  any  constant  can 
be  made  less  than  any  assigned  value. 

Given  a  variable  V,  which  can  be  made  less  than  any  previ- 
ously assigned  value,  however  small,  and  let  K  be  any  constant. 

To  prove  that  v  •  K  may  be  made  as  small  as  we  please,  i.e. 
less  than  any  assigned  value. 

Assign  any  value,  as  a,  no  matter  how  small. 

Now  a  value  for  v  may  be  found  as  small  as  we  please. 

Take  V  <  — .    Then  v  •  K  <  a ;  i.e.  V  •  K  may  be  made  less 
K 

than  any  assigned  value.  Q.E.D. 

586.  Cor.  I.     If  a  variable  can  be  made  less  than  any 
assigned  value,  the  quotient  of  the  variable  by  any  con- 
stant, except  zero,  can  be  made  less  than  any  assigned 
value. 

HINT.     —  -  —  •  F,  which  is  the  product  of  the  variable  and  a  constant. 
K     K 

587.  Cor.  II.     If  a  variable  can  be  made  less  than  any 
assigned  value,  the  product  of  that  variable  and  a  de- 


292  PLANE   GEOMETRY 

creasing   value   may  be   made  less   than   any  assigned 
value. 

HINT.     Apply  the  preceding  theorem,  using  as  K  a-  value  greater  than 
any  value  of  the  decreasing  multiplier. 

588.  Cor.  m.   The  product  of  a  variable  and  a  variable 
may  be  a  constant  or  a  variable. 

589.  Cor.  IV.    If  a  variable  can  be  made  less  than  any 
assigned  value,  the  square  of  that  variable  can  be  made 
less  than  any  assigned  value.    (Apply  Cor.  II.) 


Ex.  1151.  Which  of  the  corollaries  under  Prop.  I  is  illustrated  by  the 
theorem:  "  The  product  of  the  segments  of  a  chord  drawn  through  a  fixed 
point  within  a  circle  is  constant ' '  ? 

PROPOSITION  II.     THEOREM 

590.  The  limit  of  the  product  of  a  variable  and  a  con- 
stant, not  zero,  is  the  limit  of  the  variable  multiplied 
by  the  constant. 

Given   any  variable  F  which,  approaches  the  finite  limit  L, 
and  let  K  be  any  constant  not  zero. 
To  prove   the  limit  of  K  •  V=  K  -  L. 
Let#  =  £  —  F;  then  v  =  L  —  R. 
.'.  K-  V=K>  L  —  K  -  R. 
But  the  limit  of  K  -  R  =  0. 
.-.  the  limit  of  K  •  v—  the  limit  of  (K  -  L  —  K  •  R )  =  K  •  L. 

Q.E.D. 

591.  Cor.     The  limit  of  the  quotient  of  a  variable  by 
a  constant  is  the  limit  of  the   variable   divided  by  the 
constant. 

HINT.     —  =  —  •  "F,  which  is  the  product  of  the  variable  and  a  constant. 
K     K 

PROPOSITION  III.     THEOREM 

592.  If  two  variables  approach  finite  limits,  not  zero, 
then  the  limit  of  their  product  is  equal  to  tJie  product 
of  their  limits. 


APPENDIX  293 

Given  variables  F  and  F'  which  approach  the  finite  limits  L 
and  L',  respectively. 

To  prove    the  limit  of  F  •  F'  =  L  •  L1. 

Let  R  =  L  —  F  and  X1  =  L'  —  F'. 
Then  V—  L  —  E  and  F'  =  £'  —  R'. 
.*.  F  •  F'  =  L  •  L'  —  (L1  •  R  -f  i  .  72'  —  £  .  R1). 
But  the  limit  of  (L1  •  R  +  £  -  R1  —  R  -  R')=  0. 
.-.  the  limit  of  F  •  F'  =  the  limit  of  \_L  •  L'  —  (L1  -  R  -f  L  •  R'  — 
R-  £')]  =  £  •£'• 

.-.  the  limit  of  F  -  F'  =  L  -  L'.  Q.E.D. 

593.  Cor.     If  each   of  any  finite  number  of  variables 
approaches  a  finite  limit,  not  zero,   then  the   limit  of 
their  product  is  equal  to  the  product  of  their  limits. 

PROPOSITION  IV.     THEOREM 

594.  If  two  related  variables   are  such  that   one  is 
always  greater  than  the  other,  and  if  the  greater  con-? 
tinually  decreases  while  the  less  continually  increases, 
so  that  the  difference  between  the  two  may  be  made  as 
small  as  we  please,  then  the  two  variables  have  a  com- 
mon limit  which  lies  between  them. 


A  P   Q   R       L      R'  Q'  P' 

Given  the  two  related  variables  AP  and  AP',  AP'  greater  than 
AP,  and  let  AP  and  AP1  be  such  that  as  AP  increases  AP1  shall 
decrease,  so  that  the  difference  between  AP  and  AP'  shall  ap- 
proach zero  as  a  limit. 

To  prove  that  AP  and  AP1  have  a  common  limit,  as  AL,  which 
lies  between  AP  and  AP'. 

Denote  successive  values  of  AP  by  AQ,  AR,  etc.,  and  denote 
the  corresponding  values  of  AP'  by  AQ\  AR1,  etc. 

Since  every  value  which  AP  assumes  is  less  than  any  value 
which  AP'  assumes  (Hyp.)     .-.  AP  <  AR', 
^  i§  continually  increasing, 


294  PLANE   GEOMETRY 


A  P   Q    R       L       R'  Q'  P' 

Hence  AP  has  some  Jimit.     (By  def.  of  a  limit,  §  349.) 

Since  any  value  which  APf  assumes  is  greater  than  every 
value  which  AP  assumes  (Hyp.)  .'.  AP'  >  AR. 

But  AP'  is  continually  decreasing. 

Hence  AP'  has  some  limit.     (By  def.  of  a  limit,  §  349.) 

Suppose  the  limit  of  AP  =£  the  limit  of  AP1. 

Then  let  the  limit  of  AP  be  AK,  while  that  of  AP'  is  AK1. 
Then  AK  and  AK'  have  some  finite  values,  as  m  and  m',  and 
their  difference  is  a  finite  value,  as  d. 

But  the  difference  between  some  value  of  AP  and  the  cor- 
responding value  of  AP1  cannot  be  less  than  the  difference  of 
the  two  limits  AK  and  AK'. 

This  contradicts  the  hypothesis  that  the  difference  between 
AP  and  AP1  shall  approach  zero  as  a  limit. 

.•.  the  limit  of  AP  =the  limit  of  AP1  and  lies  between  AP 
•and  AP',  as  AL.  Q.E.D. 

595.  THEOREM.  With  every  straight  line  segment  there 
is  associated  a  number  which  may  be  called  its  measure- 
number. 

a 


For  line  segments  commensurable  with  the  unit  this  theo- 
rem was  considered  in  §§  335  and  336;  we  shall  now  consider 
the  case  where  the  segment  is  incommensurable  with  the 
chosen  unit. 

Given  the  straight  line  segment  a  and  the  unit  segment  u; 
to  express  a  in  terms  of  u. 

Apply  u  (as  a  measure)  to  a  as  many  times  as  possible,  sup- 
pose t  times,  then 

t  •  u  <  a  <  (t  + 1)  u. 

Now  apply  some  fractional  part  of  u,  say  -,  to  a,  and  sup- 

P 


APPENDIX  295 

pose  it  is  contained  ^  times,  then 


p  p 

Then  apply  smaller  and  smaller  fractional  parts  of  u  to  a,  say 
—  ,  —  ,  —  ,  •••,  and  suppose  them  to  be  contained  t>,  £3,  t4,  •  ••  times 

respectively,  then 


p2  [)  P  P 

Now  the  infinite  series  of  increasing  numbers  t,  J,  —  ,  ••-, 

p  p* 

none  of  which  exceeds  the  finite  number  t  +  1,  defines  a  num- 
ber n  (the  limit  of  this  series)  which  we  shall  call  the  measure- 
number  of  a  with  respect  to  u.  Moreover,  this  number  n  is 
unique,  i.e.  independent  of  p  (the  number  of  parts  into  which 
the  unit  was  divided),  for  if  m  is  any  number  such  that  m  <  n, 
then  m  •  u  <  a,  and  if  m  >  n,  then  m  •  u  >  a  ;  we  are  therefore 
justified  in  associating  the  number  n  with  a,  and  in  saying 
that  n  •  u  =  a. 

596.  Note.     Manifestly,  the  above  procedure  may  be  applied  to  any 
geometric   magnitude   whatever,  i.e.  every  geometric   magnitude  has  a 
unique  measure-number. 

597.  Cor.     If  a  magnitude  is  variable  and  approaches 
a  limit,  then,  as  the  magnitude  varies,  the  successive 
measure-numbers  of  the  variable  approach  as  their  limit 
the  measure-number  of  the  limit  of  the  magnitude. 

598.  Discussion  of  the  problem  : 

To  determine  whether  two  given  lines  are  commen- 
surable or  not  ;  and  if  they  are  commensurable,  to  find 
their  common  measure  and  their  ratio  (§  345). 

Moreover,  GD  is  the  greatest  common  measure  of  AB  and  CD. 
For  every  measure  of  AB  is  a  measure  of  its  multiple  CE. 
Hence,  every  common  measure  of  AB  and  CD  is  a  common 
measure  of  CE  and  CD  and  therefore  a  measure  of  their  differ- 


I 

296  PLANE   GEOMETRY 

ence  ED,  and  therefore  of  AF,  which  is  a  multiple  of  ED. 
Hence,  every  common  measure  of  AB  and  CD  is  a  common 
measure  of  AB  and  AF  and  therefore  a  measure  of  their  differ- 
ence FB.  Again,  every  common  measure  of  ED  and  FB  is  a 
common  measure  of  ED  and  EG  (a  multiple  of  FB)  and  hence  of 
their  difference  GD.  Hence,  no  common  measure  of  AB  and  CD 
can  exceed  GD.  Therefore,  GD  is  the  greatest  common  measure 
of  AB  and  CD. 

Now,  if  AB  and  CD  are  commensurable,  the  process  must  ter- 
minate; for  any  common  measure  of  AB  and  CD  is  a  measure 
of  each  remainder,  and  every  segment  applied  as  a  measure  is 
less  than  the  preceding  remainder.  Now,  if  the  process  did 
not  terminate,  a  remainder  could  be  reached  which  would  be 
less  than  any  assigned  value,  however  small,  and  therefore  less 
•than  the  greatest  common  measure,  which  is  absurd. 

If  AB  and  CD  are  incommensurable,  the  process  will  not  ter- 
minate; for,  if  it  did,  the  last  remainder  obtained  would  be  a 
common  measure  of  AB  and  CD,  as  shown  above. 

599.  THEOREM.     An  angle  can  be  bisected  by  only  one 
line. 

Given  Z  ARC,  bisected 
by  BT. 

To  prove  that  no  other 
bisector  of  /.  ABC  exists. 

Suppose  that  another 
bisector  of  /.ABC  exists, 
e.g.  BF. 

Then  Z  ABF  —  Z.  ABT.     This  is  impossible. 

.-.  no  other  bisector  of  /.ABC  exists.  Q.E.D. 

600.  Note  on  Axioms.      The  thirteen  axioms  (§  54)  refer  to  num- 
bers and  may  be  used  when  referring  to  the  measure-numbers  of  geometric 
magnitudes.     Axioms  2-9  are  not  applicable  always  to   equal  figures. 
(See  Exs.  800  and  801.)     Axioms   7   and   8   hold  for  positive  numbers 
only,  but  do  not  hold  for  negative  numbers,  for  zero,  nor  for  infinity  ; 
axioms  11  and  12  hold  only  when  the  number  of  parts  is  finite. 


INDEX 


(The  numbers  refer  to  articles.) 


ART. 

Acute  triangle 98 

Adjacent  angles 42 

Alternation 396 

Altitude  ....       100,  228,  229 
Analysis  of  a  proportion     .     .  397 
of  constructions    .... 
151,  152,  170  (Ex.  187),  274, 
499 

of  exercises 274 

Angle 38 

acute 47 

at  center  of  circle  .  .  .  292 
at  center  of  regular  polygon  534 
bisector  of  ...  53,  127,  599 

central 292 

degree  of 71 

inscribed  in  circle  .  .  .  363 
inscribed  in  segment  .  .  364 

magnitude  of 41 

obtuse 48 

of  one  degree 71 

reflex 49 

right 45,  71 

right  and  left  side  of      .     .  199 

sides  of 38 

straight 69 

vertex  of 38 

Angles 38 

added 43 

adjacent 42 

alternate  exterior  .   182 


ART. 

Angles,  alternate  interior   .     .  182 
complementary     ....     68 

corresponding 182 

designation  of  ....    39,  50 

difference  of 44 

equal 44 

exterior 87,  182 

exterior  interior    ....  182 
homologous       .     .     .     417,  424 

interior 86,  182 

oblique 51 

of  a  polygon 8(5 

subtracted 44 

sum  of 43 

supplementary      ....     69 
supplementary-adjacent      .     72 

vertical 70 

Antecedents 384 

Apothem 533 

Application 79 

Arc 121,  280 

degree  of 296 

length  of 551 

major 291 

minor 291 

Arcs,  similar     ......  565 

Area  of  a  figure     .     .     .  •  .     .  471 

of  circle 558 

of  rectangle      .     .  ...     468-471 

of  surface 407 

Argument  only      .     .     .     .        171 


297 


298 


INDEX 


ART. 

Arms  of  isosceles  triangle  .  .  94 
of  right  triangle  ....  96 

Assumptions 12,  54 

Axiom 55 

Axioms 54 

•  note  on 600 

Rase  of  isosceles  triangle  .  .  94 
of  polygon 99 

Bisector  of  angle  .  .53,  127,  599 
of  line 52,  145 

Bisectors  of  angles  of  triangle     126 

Center-line  .     .     . "  .     .     .     .325 

Center  of  circle      .     .     .     119,278 

of  regular  polygon     .     .     .531 

Central  angle 292 

Centroid .     .205 

Chord 281 

Circle       .     .     .     .     .     .     119,  276 

arc  of 121,  280 

area  of 558 

center  of      ....     119,  278 

chord  of       281 

circumference  of  .  119,  277,  550 

circumscribed 300 

determined 324 

diameter  of       .'....  282 

escribed 322 

inscribed 317 

radius  of       ....     119,  278 

secant  of      .......  285 

sector  of       ....     287,  564 

segment  of 288 

tangent  to 286 

Circles,  concentric      ....  326 

tangent,  externally    .     .     .  330 

tangent,  internally    .     .     .  330 

Circumference  .     .     .119,  277,  550 

length  of      .     .     .     .     550,  552 

measurement  of    ....  540 

Circumscribed  circle  ....  300 


Circumscribed  polygon  . 
Clockwise  motion  .  .  . 
Closed  figure  .... 

Closed  line 

Coincidence       .... 
Collinear  segments     .     . 
Commensurable  quantities 
337, 

Common  measure 
Compasses,  use  of 
Complementary  angles  . 
Complete  demonstration 
Composition      .... 
Composition  and  division 
Concentric  circles       .     . 
Conclusion  .     .     .     .     . 
Concurrent  lines    .     .     . 
Consequents      .... 

Constant 

Construction     .... 

of  triangles  .... 

Continued  proportion     . 

Converse  theorem       .     . 

Corollary 

Counter-clockwise  motion 


ART. 

317 
86 
83 
82 
17 
28 


342,  469 
.  .  337 
.  .  31 
.  .  68 
.  .  171 
.  .  398 
.  .  400 
.  .  326 
.  .  57 
.  .  190 
.  .  334 
346,  347 
.  .  123 
.  .  269 
.  .  405 
.  .  135 
.  .  58 
50 


Definition,  test  of  ....  138 
Degree,  of  angle 71 

of  arc 296 

Demonstration 56,  79 

complete 171 

Determined  circle  ....  324 
Determined  line  and  point  .  25,  26 
Determined  triangle  .  .  133,  209 

Diagonal 88 

Diameter .  282 

Dimensions 10 

Discussion 123 

Distance  between  two  points  .  128 

from  point  to  line  .  .  .166 

Division .  399 

Drawing  to  scale 437 


INDEX 


299 


ART. 

Equal  figures 18,  473 

homologous  parts  of  .     109,  110 
Equiangular  polygon       .    *.     .     90 

Kquidistance 128 

Equilateral  polygon  ....  89 
Equilateral  triangle  ....  95 
Equivalent  figures  ....  473 

Escribed  circle 322 

Euclid's  Elements  ....  114 
Exterior  angle  of  polygon  .  .  87 
Extreme  and  mean  ratio 

464,  465  (Ex.  763),  526 
Extremes 385 

Figure,  closed 83 

geometric 13 

plane 36 

rectilinear 37 

Figures,  equal 18,  473 

equivalent 473 

isoperimetric 572 

similar 419,  473 

transformation  of  ...  493 

Formulas  of  plane  geometry    .  570 

Fourth  proportional  ....  387 

Geometric  figure 13 

Geometric  figures,  equal  .  .  18 
Geometric  solid  .  .  .  2,  9,  11 

Geometry 14 

plane 36 

subject  matter  of  ....  1 

Golden  section 464 

Harmonical  division  ....  434 

Hexagon 92 

Historical  notes 

Achilles  and  the  tortoise     .  354 

Ahmes 474,  569 

Archimedes ....  542,  569 
Area  of  a  circle  ....  569 
Cicero  .  .  542 


ART. 

Descartes 237 

Division  of  circumference  .  516 
Egyptians     .,...-. 

371,  449,  474,  497,  569 
Euclid     .     .     .     .114,  510,  516 

Gauss 516,  520 

Herodotus    .     .     .     .     .     .  474 

Hero  of  Alexandria  .     .     .  569 

King  Hiero 542 

Lambert 569  • 

Lindemann 569 

Metius  of  Holland     .     .     .  569 

Morse,  S.  F.  B 520 

Newton 542 

Origin  of  geometry    ...  474 
Plato        .     .     .     .     .     268,  542 

Plutarch 371 

Pons  asinorum 114 

Ptolemy 569 

Pythagoras  .     .       344,  510,  514 

Richter   . 569 

Seven  Wise  Men   .     .     .     .371 

Shanks .  569 

Squaring  the  circle    .     .     .569 

Stobseus 114 

Thales 371,  510 

Zeno   .     .     . v  .     .     .     .     .354 

Homologous  parts 

109,  110,  417,  418,  424 

Hypotenuse 96 

Hypothesis   .......    57 

Incommensurable  quantities    . 

339,  343,  344,  470 
Indirect  proof   ....     159,  161 

Inscribed  angle 363 

Inscribed  circle 317 

Inscribed  polygon 299 

Instruments,  use  of    ....     31 

Inversion 395 

Isoperimetric  figures 572 

Isosceles  trapezoid      ....  227 


300 


INDEX 


ART 

Isosceles  triangle 94 

arms  of 94 

base  of 94 

sides  of 94 

vertex  angle  of       ....    94 

Length  of  arc "551 

of  circumference    .     .     550,  552 
Length  of  perpendicular     .     .  166 

of  secant 426 

of  tangent 319 

Limits       .      346,  349^351,  590-594 

Line 4,  7,  11,  27 

bisected 52,  145 

broken 30 

closed 82 

curved      .......     29 

determined 25 

divided  externally      .     .     .  406 
divided  harmonically      .     .  434 
divided    in    extreme    and 
mean  ratio       .     ... 

.   464,  465  (Ex.  763),  526 
divided  internally      .     .     .  406 

of  centers 325 

right 23 

segments  of  .  /.     .     .       27,406 

straight 21,  23 

Line  segment     ......     27 

Lines 20 

concurrent 196 

difference  of 31 

divided  proportionally    .     .  407 
homologous     109,  110,  418,  424 

oblique 46 

parallel 177 

perpendicular 45 

product  of  .  425,  475,  476,  511 
rectangle  of  .  425,  475,  476,  611 
Locus  .....  129,  130,  131 
as  an  assemblage  ....  130 
as  a  path 130 


ART. 
Locus,  finding  of  a     ....  144 

of  a  point 130 

of  alf  points 130 

problem,  solution  of  .     .     .  143 

Magnitude  of  an  angle    .     .     .    41 

Major  arc 291 

Maximum 571 

Mean  proportional 386 

Means 385 

Measure,  commoi^ 337 

numerical      ,  V.     .    .    335,  467 

unit  of 335,  466 

Measure-number,  335,  467,  471,  595 

Measurement 335 

of  angle     .......  358 

of  arc 358 

of  circle  .  ,  .  .  .  540,  558 
of  circumference  .  .  540,  552 
of  distances  by  means  of 

triangles 115 

of  line. 335,  595 

of  rectangle.     .     .     .     468-471 

of  surface 466 

Median,  of  trapezoid  ....  25t 

of  triangle 102 

Median  center  of  triangle    .     .  265 

Methods  of  attack 

103,  104,  106,  110  (Ex.  62), 
115,  151,  152,  158,  159,  176 
(Ex.  187),  269-274,  275,  397, 
398,  425,  499,  502,  513. 

Minimum .^.^11 

Minor  arc  .     .     .     .291 


Numerical  measure 


.  335,  467 


Oblique  angles 51 

Oblique  lines 46 

Obtuse  triangle       97 

One  to  one  correspondence  595,  596 
Opposite  theorem.  ,    %     .    •    •  136 


INDEX 


301 


ART. 

Optical  illusions 81 

Outline  of  proof 171 

Parallel  lines    ' 177 

Parallelogram 220 

altitude  of 228 

base  of 228 

Parallelograms  classified     223,  243 

Pentagon 92 

Perimeter  of  polygon       ...     85 
Perpendicular    ....      45,  166 

Pi .  554 

evaluation  of 568 

Plane 32,  33,  34 

Plane  figure .    36 

Plane  geometry 36 

Point 5,  6,  11 

determined 26 

of  tangency 286 

Polygon 84 

angles  of 86 

base  of .     00 

circumscribed 317 

diagonal  of  .     .     . "    .     .     .     88 
equiangular       ......     90 

'    equilateral    .     .     . »  .     .     .     89 
exterior  angles  of  ....     87 

inscribed 299 

perimeter  of 85 

regular 91,  515 

sides  of 84 

vertices  of 84 

Polygons, mutually  equiangular  417 
sides  proportional       .     .     .418 

similar 419 

Pons  asinorum 114 

Portraits 

Archimedes 542 

Descartes 237 

Euclid 114 

Gauss 520 

Pythagoras 510 


,  ART. 

Thales 371 

Postulate 15 

circle 122 

parallel  line      .     .     .     178,  179 

revolution 40,  54 

straight  line  .  .  .  .  24,  54 
transference  .  .  .  .  16,  64 

Problem  defined 59 

Problems  of  computation     .     .     59 
Problems  of  construction     .     .     59 

analysis  of 

151,    152,    176  (Ex. 
187),  274,  499 

discussion  of 123 

proof  of  . 123 

solution  of 123 

Product  of  lines   425,  475,  476,  511 

Projection  of  line 451 

of  point 450 

Proof 79,  123 

analytic  method  ....  275 
by  exclusion  .  .  .  159,  161 
by  successive  substitutions  275 

indirect 159,  161 

necessity  for 81 

outline  of 171 

reductio  ad  absurdum  159,  161 
synthetic  method  .  .  .  275 

Proportion 382 

analysis  of 397 

antecedents  of 384 

by  alternation 396 

by  composition  .  .  .  398 
by  composition  and  division  400 

by  division        399 

by  inversion  .  .  .  .  .  395 
consequents  of  ....  384 

continued 405 

extremes  of 385 

means  of 385 

terms  of 383 

ways  of  writin  •     ....  382 


302 


INDEX 


ART;. 
Proportional,  fourth  ....  387 

mean 380 

third   .     .     .    • 386 

Proportions  simplified     .     380,  397 
Proposition 56 

Quadrant 295 

Quadrilateral     ......     92 

Quadrilaterals  classified       220,  243 


Radius  of  circle      .     . 
of  regular  polygon 


119,  278 
.     .  532 
Ratio    ....    340-343,  382,  384 

antecedent  of 384 

consequent  of 384 

extreme  and  mean     .     .     . 

464,  465  (Ex.  763),  526 
of  any  two  surfaces   .     .     .  472 

of  similitude 418 

of  two  commensurables  .  342 
pf  two  incommensurables  .  343 
of  two  magnitudes  .  .  .  341 

Rectangle 223 

area  of 468-471 

of  two  lines     425,  475,  476.  511 

Rectilinear  figure        ....     37 

Reductio  ad  absurdum    .     159,  161 

Regular  polygon    .     .     .       91,  515 

angle  at  center  of       ...  534 

apothem  of       533 

center  of 531 

radius  of 532 

Related  variables  .     .     .     352,  594 

Rhomboid     . 224 

Rhombus      .......  226 

Right  triangle 95 

arms  of 96 

hypotenuse  of        ....     96 

sides  of    .......     96 

Ruler,  use  of 31 


Scale,  drawing  to 


.  437 


ART. 

Scalene  triangle 93 

Secant 285 

length  of 426 

Sector      ......     287,  564 

Segment,  of  circle       ....  288 

of  line 27 

Segments,  added 31 

collinear 28 

difference  of 31 

of  a  line 406 

similar     ........  565 

subtracted 31 

sum  of 31 

Semicircle 289 

Semicircumference     ....  290 
Series  of  equal  ratios      .     401,  405 

Sides,  of  angle. 38 

of  polygon 81 

Similar  arcs 5(55 

Similar  polygons    .     .     .     .     .419 

Similar  sectors 565 

Similar  segments  ....  565 
Similarity  of  triangles  .  .  .  431 
Similitude,  ratio  of  ....  418 
Solid,  geometric  .  .  .  2,  9,  11 
Solution  of  exercises  .  .  .  275 
of  locus  problems  .  .  .  143 

of  problems 123 

of  theorems 79 

Square 225 

Squaring  the  circle     ....  569 

Straight  angle 69 

Straightedge 21 

Straight  line 21,  23 

determined 25 

Summary,  of  divisions  of  cir- 
cumference    ' .     .     .     .  529 
of  equal  triangle  theorems     118 
of  formulas  of  plane  geom- 
etry     570 

of  parallel  line  theorems     .  197 
of  quadrilaterals   ....  243 


INDEX  . 


303 


Summary,    of    similar   triangle 

theorems 431 

of  trapezoids     .      Exs.  320,  321 

of  unequal  angle  theorems     174 

of  unequal  line  theorem.8    .   175 

Superposition    .     .     .    lit,  103,  106 

Supplementary  angles     ...     69 

Supplementary -adjacent  angles    72 

Surface     .     .     .     .     .     .      3,  8,  11 

curved 35 

plane       ....       32,  33,  34 
Surfaces        32 

Tangent        286 

external  common  ....  333 

internal  common  ...-.•  333 

length  of 319 

Theorem       57 

converse  of 135 

opposite  of 136 

solution  of   .     .     .     .     .     .  79 

Third  proportional      ....  386 

Transformation 493 

Transversal  .  • 181 

Trapezium 222 

Trapezoid 221 

altitude  of 229 

bases  of 229 

isosceles 227 

median  of 251 

Triangle 92 

acute  .  98 


ART 

Triangle,  altitudes  of  ...  100 
bisectors  of  angles  of  .  .126 

•centroid  of 265 

construction  of  ....  269 
determined  .  .  .  .  133,  269 

equilateral 95 

isosceles       94 

median  center  of  ....  265 

medians  of 102 

notation  of 270 

obtuse      .......     97 

parts  of 269 

right .96 

scalene     .......     93 

vertex  of 99 

vertex  angle  of  '   .     .     .   94,  99 

Triangles,  classified   .     .     .    93-98 

similarity   of 431 

Unit  of  measure     .     .     .     335.  466 

Variable  .  .  346,  348,  585-594 
approaches  its  limit  .  .  .  350 

independent 346 

limit  of 349 

reaches  its  limit  ....  351 
related  .....  352,  594 

Vertex  of  angle  .  .  .  .  .  38 

of  polygon 84 

of  triangle 99 

Vertex  angle  of  triangle      .    94,  99 

Vertical  angles 70 


ELEMENTS      OF 
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By  ANDREW  W.   PHILLIPS,  Ph.D.,  and  WENDELL 
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are  set  forth  briefly  and  clearly  the  scope  and  leading  charac- 
teristics of  each  of  our  best  text-books.      In  most  cases  there 
are  also  given  testimonials  from  well-known  teachers,  which 
have  been  selected  quite  as  much  for  their  descriptive  qualities 
as  for  their  value  as  commendations. 

^|  For  the  convenience  of  teachers  this  Catalogue  is  also 
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study.  These  pamphlets  are  entitled  :  English,  Mathematics, 
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^[  Teachers  seeking  the  newest  and  best  books  for  their 
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^|  Copies  of  our  price  lists,  or  of  special  circulars,  in  which 
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LIBKAPV 

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Berkeley 


THE  UNIVERSITY  OF  CALIFORNIA 


YB   17313 

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